Chapter 7
Exponential and Logarithmic Functions
Section 7.1
12. a.
1. 43 = 64 ; 82 / 3 =
( 8)
3
2
= 22 = 4 ; 3−2 =
1
32
=
1
9
2. 5 x − 2 = 3
5x = 5
x =1
The solution set is {1} .
3. False. To obtain the graph of y = ( x − 2)3 , we
would shift the graph of y = x3 to the right 2 units.
4.
f (4) − f (0) [3(4) − 5] − [3(0) − 5]
=
4−0
4
(12 − 5) − (0 − 5)
=
4
7 − (−5)
=
4
12
=
4
=3
51.7 ≈ 15.426
b.
51.73 ≈ 16.189
c.
51.732 ≈ 16.241
d.
5
3
≈ 16.242
13. a.
23.14 ≈ 8.815
b.
23.141 ≈ 8.821
c.
23.1415 ≈ 8.824
d.
2π ≈ 8.825
14. a.
22.7 ≈ 6.498
b.
22.71 ≈ 6.543
c.
22.718 ≈ 6.580
d.
2e ≈ 6.581
15. a.
3.12.7 ≈ 21.217
b.
3.142.71 ≈ 22.217
5. True
c.
3.1412.718 ≈ 22.440
1⎞
⎛
6. ⎜ −1, ⎟ , ( 0,1) , (1, a )
a⎠
⎝
d.
πe ≈ 22.459
16. a.
7. 1
8. 4
9. False
10. False. The range will be {x | x > 0} or (0, ∞) .
11. a.
32.23 ≈ 11.587
c.
32.236 ≈ 11.664
d.
3
5
b.
2.713.14 ≈ 22.884
c.
2.7183.141 ≈ 23.119
d.
e π ≈ 23.141
17. e1.2 ≈ 3.320
32.2 ≈ 11.212
b.
2.73.1 ≈ 21.738
18. e−1.3 ≈ 0.273
19. e−0.85 ≈ 0.427
20. e2.1 ≈ 8.166
≈ 11.665
616
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
21.
x
y = f ( x)
–1
3
0
6
1
12
2
18
3
f ( x + 1)
f ( x)
24.
6
=2
3
12
=2
6
18 3
=
12 2
22.
–1
2
3
0
1
1
3
2
3
1
2
(9 / 4) 9 2 3
= ⋅ =
(3 / 2) 4 3 2
2
9
4
( 27 / 8) 27 4 3
=
⋅ =
(9 / 4) 8 9 2
3
27
8
x
y = g ( x)
–1
2
0
5
1
8
2
11
3
14
g ( x + 1)
g ( x)
5
2
8
5
x
y = H ( x)
H ( x + 1)
H ( x)
–1
1
4
1
=4
(1/ 4 )
0
1
1
4
2
16
3
64
F ( x)
1
3 3
= 1⋅ =
2
/
3
2 2
( )
(3 / 2)
=
Yes, an exponential function since the ratio of
3
consecutive terms is constant with a = . So the
2
3
base is .
2
25.
Not an exponential function since the ratio of
consecutive terms is not constant.
23.
y = F ( x)
30
Not an exponential function since the ratio of
consecutive terms is not constant.
F ( x + 1)
x
x
y = f ( x)
–1
3
2
0
3
1
6
2
12
3
24
f ( x + 1)
f ( x)
3
2
= 3⋅ = 2
3
(3 / 2)
6
=2
3
12
=2
6
24
=2
12
Yes, an exponential function since the ratio of
consecutive terms is constant with a = 2 . So the
base is 2.
4
=4
1
16
=4
4
64
=4
16
Yes, an exponential function since the ratio of
consecutive terms is constant with a = 4 . So the
base is 4.
617
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
26.
x
y = g ( x)
g ( x + 1)
29. B
g ( x)
–1
6
1
6
0
1
0
=0
1
1
0
2
3
3
10
30. F
31. D
32. H
33. A
34. C
35. E
36. G
Not an exponential function since the ratio of
consecutive terms is not constant.
27.
x
y = H ( x)
–1
2
0
4
1
6
2
8
3
10
37.
H ( x + 1)
f ( x) = 2 x + 1
Using the graph of y = 2 x , shift the graph up 1
unit.
Domain: All real numbers
Range: { y | y > 0} or (1, ∞)
Horizontal Asymptote: y = 1
H ( x)
4
=2
2
6 3
=
4 2
Not an exponential function since the ratio of
consecutive terms is not constant.
28.
x
y = f ( x)
–1
1
2
0
1
4
1
1
8
2
1
16
3
1
32
f ( x + 1)
f ( x)
38.
(1/ 4 ) 1 2 1
= ⋅ =
(1/ 2 ) 4 1 2
(1/ 8) 1 4 1
= ⋅ =
(1/ 4 ) 8 1 2
(1/16 ) 1 8 1
= ⋅ =
(1/ 8) 16 1 2
(1/ 32 ) 1 16 1
= ⋅ =
(1/16 ) 32 1 2
f ( x ) = 3x − 2
Using the graph of y = 3x , shift the graph down 2
units.
Domain: All real numbers
Range: { y | y > −2} or ( −2, ∞ )
Horizontal Asymptote: y = −2
Yes, an exponential function since the ratio of
1
consecutive terms is constant with a = . So The
2
1
base is .
2
618
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
39.
f ( x ) = 3x −1
Using the graph of y = 3x , shift the graph right 1
unit.
Domain: All real numbers
Range: { y | y > 0} or ( 0, ∞ )
Horizontal Asymptote: y = 0
42.
⎛1⎞
f ( x) = 4 ⋅⎜ ⎟
⎝3⎠
x
x
40.
⎛1⎞
Using the graph of y = ⎜ ⎟ , vertically stretch the
⎝3⎠
graph by a factor of 4. That is, for each point on the
graph, multiply the y-coordinate by 4.
Domain: All real numbers
Range: { y | y > 0} or ( 0, ∞ )
f ( x) = 2 x + 2
Using the graph of y = 2 x , shift the graph left 2
units.
Domain: All real numbers
Range: { y | y > 0} or (0, ∞)
Horizontal Asymptote: y = 0
Horizontal Asymptote: y = 0
43.
41.
⎛1⎞
f ( x) = 3⋅⎜ ⎟
⎝2⎠
f ( x) = 3− x − 2
Using the graph of y = 3x , reflect the graph about
the y-axis, and shift down 2 units.
Domain: All real numbers
Range: { y | y > −2} or (− 2, ∞)
Horizontal Asymptote: y = − 2
x
x
⎛1⎞
Using the graph of y = ⎜ ⎟ , vertically stretch the
⎝2⎠
graph by a factor of 3. That is, for each point on the
graph, multiply the y-coordinate by 3.
Domain: All real numbers
Range: { y | y > 0} or ( 0, ∞ )
Horizontal Asymptote: y = 0
619
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
44.
f ( x) = −3x + 1
Using the graph of y = 3x , reflect the graph about
the x-axis, and shift up 1 unit.
Domain: All real numbers
Range: { y | y < 1} or (− ∞, 1)
Horizontal Asymptote: y = 1
47.
f ( x ) = 2 + 3x / 2
Using the graph of y = 3x , stretch the graph
horizontally by a factor of 2, and shift up 2 units.
Domain: All real numbers
Range: { y | y > 2} or (2, ∞)
Horizontal Asymptote: y = 2
45.
f ( x) = 2 + 4 x −1
Using the graph of y = 4 x , shift the graph to the
right one unit and up 2 units.
Domain: All real numbers
Range: { y | y > 2} or (2, ∞)
Horizontal Asymptote: y = 2
48.
46.
f ( x) = 1 − 2− x / 3
Using the graph of y = 2 x , stretch the graph
horizontally by a factor of 3, reflect about the yaxis, reflect about the x-axis, and shift up 1 unit.
Domain: All real numbers
Range: { y | y < 1} or (−∞, 1)
Horizontal Asymptote: y = 1
f ( x) = 1 − 2 x +3
Using the graph of y = 2 x , shift the graph to the
left 3 units, reflect about the x-axis, and shift up 1
unit.
Domain: All real numbers
Range: { y | y < 1} or (−∞, 1)
Horizontal Asymptote: y = 1
620
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
49.
f ( x) = e − x
Using the graph of y = e x , reflect the graph about
the y-axis.
Domain: All real numbers
Range: { y | y > 0} or (0, ∞)
Horizontal Asymptote: y = 0
52.
f ( x) = e x − 1
Using the graph of y = e x , shift the graph down 1
unit.
Domain: All real numbers
Range: { y | y > −1} or (−1, ∞)
Horizontal Asymptote: y = −1
50.
f ( x ) = −e x
Using the graph of y = e x , reflect the graph about
the x-axis.
Domain: All real numbers
Range: { y | y < 0} or (−∞, 0)
Horizontal Asymptote: y = 0
53.
51.
f ( x) = 5 − e− x
Using the graph of y = e x , reflect the graph about
the y-axis, reflect about the x-axis, and shift up 5
units.
Domain: All real numbers
Range: { y | y < 5} or (−∞, 5)
Horizontal Asymptote: y = 5
f ( x) = e x + 2
Using the graph of y = e x , shift the graph 2 units
to the left.
Domain: All real numbers
Range: { y | y > 0} or (0, ∞)
Horizontal Asymptote: y = 0
621
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
54.
y
f ( x) = 9 − 3e − x
8
Using the graph of y = e x , reflect the graph about
the y-axis, stretch vertically by a factor of 3, reflect
about the x-axis, and shift up 9 units.
Domain: All real numbers
Range: { y | y < 9} or (−∞, 9)
Horizontal Asymptote: y = 9
y=7
(0, 4)
−5
−2
5
x
57. 7 x = 73
We have a single term with the same base on both
sides of the equation. Therefore, we can set the
exponents equal to each other: x = 3 .
The solution set is {3} .
55.
58. 5 x = 5−6
We have a single term with the same base on both
sides of the equation. Therefore, we can set the
exponents equal to each other: x = −6 .
The solution set is {−6} .
f ( x) = 2 − e− x / 2
Using the graph of y = e x , reflect the graph about
the y-axis, stretch horizontally by a factor of 2,
reflect about the x-axis, and shift up 2 units.
Domain: All real numbers
Range: { y | y < 2} or (−∞, 2)
Horizontal Asymptote: y = 2
59. 2− x = 16
2− x = 24
−x = 4
x = −4
The solution set is {−4} .
60. 3− x = 81
3− x = 34
−x = 4
x = −4
The solution set is {−4} .
x
56.
f ( x) = 7 − 3e
1
⎛1⎞
61. ⎜ ⎟ =
25
⎝5⎠
−2 x
Using the graph of y = e x , reflect the graph about
x
1
⎛1⎞
⎜ ⎟ = 2
5
⎝5⎠
1
,
2
stretch vertically by a factor of 3, reflect about the
x-axis, and shift up 7 units.
Domain: All real numbers
Range: { y | y < 7} or (−∞, 7)
Horizontal Asymptote: y = 7
the y-axis, shrink horizontally by a factor of
x
2
⎛1⎞
⎛1⎞
⎜ ⎟ =⎜ ⎟
⎝5⎠
⎝5⎠
x=2
The solution set is {2} .
622
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
x
1
⎛1⎞
62. ⎜ ⎟ =
64
⎝4⎠
2
2 x
(2 )
x
1
⎛1⎞
⎜ ⎟ = 3
4
⎝4⎠
x
2
2 x2 = x
3
2 x2 − x = 0
x ( 2 x − 1) = 0
x = 0 or 2 x − 1 = 0
2x = 1
1
x=
2
⎧ 1⎫
The solution set is ⎨0, ⎬ .
⎩ 2⎭
22 x −1 = 4
22 x −1 = 22
2x −1 = 2
2x = 3
3
x=
2
67.
3 − x +14
1
5
5 x + 3 = 5 −1
x + 3 = −1
x = −4
The solution set is {−4} .
68.
2 − x +15
3x = 9 x
3
( )
x3
= 32 x
x3 = 2 x
(
( )
= 33
x
3−2 x + 30 = 33 x
−2 x + 30 = 3x
30 = 5 x
6=x
The solution set is {6} .
2 x
= 3
x
9− x +15 = 27 x
(3 )
3
3
( )
= 24
2−3 x + 42 = 24 x
−3 x + 42 = 4 x
42 = 7 x
6=x
The solution set is {6} .
64. 5 x + 3 =
x3
8− x +14 = 16 x
(2 )
⎧3⎫
The solution set is ⎨ ⎬ .
⎩2⎭
65.
= 2x
22 x = 2 x
⎛1⎞
⎛1⎞
⎜ ⎟ =⎜ ⎟
⎝4⎠
⎝4⎠
x=3
The solution set is {3} .
63.
2
4x = 2x
66.
x3 − 2 x = 0
)
x x2 − 2 = 0
69.
x = 0 or x 2 − 2 = 0
x2 = 2
x=± 2
{
3x
2
−7
= 27 2 x
3x
2
−7
= 33
3x
2
−7
= 36 x
( )
2x
x2 − 7 = 6 x
}
The solution set is − 2, 0, 2 .
x2 − 6 x − 7 = 0
( x − 7 )( x + 1) = 0
x − 7 = 0 or x + 1 = 0
x=7
x = −1
The solution set is {−1, 7} .
623
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
70.
5x
2
+8
= 1252 x
5x
2
+8
= ( 53 )
5x
2
+8
= 56 x
74. e3 x = e2 − x
3x = 2 − x
4x = 2
1
x=
2
2x
x2 + 8 = 6 x
⎧1 ⎫
The solution set is ⎨ ⎬ .
⎩2⎭
x2 − 6 x + 8 = 0
( x − 4 )( x − 2 ) = 0
x − 4 = 0 or x − 2 = 0
x=4
x=2
The solution set is {2, 4} .
71.
e x = e3 x ⋅ e−2
2
2
2 x
2x
2 ⋅2
2
( )
2
⋅ 2 x = 24
x2
2 x + x2
e x = e3 x − 2
x 2 = 3x − 2
2
x 2 − 3x + 2 = 0
( x − 2 )( x − 1) = 0
8
=2
= 28
x − 2 = 0 or x − 1 = 0
x=2
x =1
The solution set is {1, 2} .
x2 + 2 x = 8
x2 + 2 x − 8 = 0
( x + 4 )( x − 2 ) = 0
x + 4 = 0 or x − 2 = 0
x = −4
x=2
The solution set is {−4, 2} .
72.
4 x
2
3 x
2
2
e 4 x + x = e12
x 2 + 4 x = 12
x 2 + 4 x − 12 = 0
( x + 6 )( x − 2 ) = 0
−1
=3
2
34 x ⋅ 33 x = 3−1
4 x +3 x2
3
2
⋅ e x = e12
e 4 x ⋅ e x = e12
2
(3 ) ⋅ (3 )
x + 6 = 0 or x − 2 = 0
x = −6
x=2
The solution set is {−6, 2} .
−1
=3
2
3 x + 4 x = −1
2
3x + 4 x + 1 = 0
( 3x + 1)( x + 1) = 0
77. a.
f ( 4 ) = 24 = 16
The point ( 4,16 ) is on the graph of f.
3x + 1 = 0
or x + 1 = 0
x = −1
3 x = −1
1
x=−
3
1⎫
⎧
The solution set is ⎨−1, − ⎬ .
3⎭
⎩
73.
(e )
76.
92 x ⋅ 27 x = 3−1
2 2x
e2
2
4 x ⋅ 2 x = 162
(2 )
1
2
e x = e3 x ⋅
75.
b.
e x = e3 x + 8
x = 3x + 8
−2 x = 8
x = −4
The solution set is {−4} .
1
16
1
2x =
16
1
x
2 = 4
2
2 x = 2−4
x = −4
1⎞
⎛
The point ⎜ −4, ⎟ is on the graph of f.
16
⎝
⎠
f ( x) =
624
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
78. a.
f ( 4 ) = 34 = 81
−2
81. a.
The point ( 4,81) is on the graph of f.
b.
79. a.
b.
1
9
1
3x =
9
1
x
3 = 2
3
3x = 3−2
x = −2
1⎞
⎛
The point ⎜ −2, ⎟ is on the graph of f.
9⎠
⎝
f ( x) =
b.
13
8
x
x
3
⎛1⎞
3⎜ ⎟ =
2
8
⎝ ⎠
x
1
⎛1⎞
⎜ ⎟ =
2
8
⎝ ⎠
x
1
⎛1⎞
⎜ ⎟ = 3
2
⎝2⎠
1
9
g ( −1) = 4 + 2 = + 2 =
4
4
9⎞
⎛
The point ⎜ −1, ⎟ is on the graph of g.
4⎠
⎝
−1
x
3
⎛1⎞
⎛1⎞
⎜ ⎟ =⎜ ⎟
⎝2⎠
⎝2⎠
x=3
13 ⎞
⎛
The point ⎜ 3, − ⎟ is on the graph of H.
8⎠
⎝
g ( x ) = 66
4 x = 64
−1
4 x = 43
x=3
The point ( 3, 66 ) is on the graph of g.
b.
H ( x) = −
13
⎛1⎞
3⎜ ⎟ − 2 = −
2
8
⎝ ⎠
4 x + 2 = 66
80. a.
2
⎛1⎞
H ( −2 ) = 3 ⎜ ⎟ − 2 = 3 ( 2 ) − 2 = 10
⎝2⎠
The point ( −2,10 ) is on the graph of H.
82. a.
b.
1
14
−3 = −
5
5
14 ⎞
⎛
The point ⎜ −1, − ⎟ is on the graph of g.
15 ⎠
⎝
g ( −1) = 5−1 − 3 =
1
⎛1⎞
F ( −1) = −2 ⎜ ⎟ + 1 = −2 ( 3) + 1 = −5
3
⎝ ⎠
The point ( −1, −5 ) is on the graph of F.
F ( x ) = −53
x
⎛1⎞
−2 ⎜ ⎟ + 1 = −53
⎝ 3⎠
x
⎛1⎞
−2 ⎜ ⎟ = −54
⎝3⎠
g ( x ) = 122
x
⎛1⎞
⎜ ⎟ = 27
⎝3⎠
3− x = 33
−x = 3
x = −3
The point ( −3, −53) is on the graph of F.
5 x − 3 = 122
5 x = 125
5 x = 53
x=3
The point ( 3,122 ) is on the graph of g.
( )
83. If 4 x = 7 , then 4 x
−2
= 7 −2
4−2 x =
4−2 x
1
72
1
=
49
625
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
( )
84. If 2 x = 3 , then 2 x
−2
2−2 x =
(2 )
2 −x
4− x
−x
85. If 3
( )
= 2 , then 3
32 x
−3
1 = k ⋅ a0
1 = k ⋅1
1= k
−2
1
22
1
=
4
p⋅ 1
and f (1) = a ( )
= 3−3
need to verify that this function yields the other
known point on the graph.
1
f ( −1) = 5−1 =
5
So we have the function f ( x ) = 5x .
53 x =
53 x
p⋅ 0
f ( 0) = k ⋅ a ( )
=2
32 x =
( )
with a > 0, a ≠ 1 . The graph contains the points
1⎞
⎛
⎜ −1, ⎟ , ( 0,1) , and (1,5 ) . In other words,
5⎠
⎝
1
f ( −1) = , f ( 0 ) = 1 , and f (1) = 5 . Therefore,
5
1
32
1
=
9
1
=
9
− x −2
86. If 5− x = 3 , then 5− x
88. We need a function of the form f ( x ) = k ⋅ a p⋅ x ,
= 3−2
5 = ap
Let’s use a = 5, p = 1. Then f ( x ) = 5x . Now we
1
33
1
=
27
89. We need a function of the form f ( x ) = k ⋅ a p⋅ x ,
87. We need a function of the form f ( x ) = k ⋅ a p⋅ x ,
with a > 0, a ≠ 1 . The graph contains the points
1⎞
⎛
⎜ −1, − ⎟ , ( 0, −1) , (1, − 6 ) , and ( 2, −36 ) . In other
6⎠
⎝
1
words, f ( −1) = − , f ( 0 ) = −1 , f (1) = − 6 , and
6
f ( 2 ) = −36 .
with a > 0, a ≠ 1 . The graph contains the points
⎛ 1⎞
⎜ −1, ⎟ , ( 0,1) , (1,3) , and ( 2,9 ) . In other words,
3⎠
⎝
1
f ( −1) = , f ( 0 ) = 1 , f (1) =3 , and f ( 2 ) = 9 .
3
p⋅ 0
Therefore, f ( 0 ) = k ⋅ a ( )
p⋅ 0
Therefore, f ( 0 ) = k ⋅ a ( )
1 = k ⋅ a0
1 = k ⋅1
1= k
−1 = k ⋅ a 0
−1 = k ⋅1
−1 = k
p⋅ 1
and f (1) = a ( )
p⋅ 1
and f (1) = −a ( )
3 = ap
Let’s use a = 3, p = 1. Then f ( x ) = 3x . Now we
−6 = −a p
6 = ap
Let’s use a = 6, p = 1. Then f ( x ) = −6 x .
need to verify that this function yields the other
1
known points on the graph. f ( −1) = 3−1 = ;
3
Now we need to verify that this function yields the
other known points on the graph.
1
f ( −1) = −6−1 = −
6
f ( 2) = 3 = 9
2
So we have the function f ( x ) = 3x .
f ( 2 ) = −62 = −36
So we have the function f ( x ) = −6 x .
626
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
90. We need a function of the form f ( x ) = k ⋅ a p⋅ x ,
92.
with a > 0, a ≠ 1 . The graph contains the points
1⎞
⎛
2
⎜ −1, − ⎟ , ( 0, −1) , (1, − e ) , and 2, −e . In other
e⎠
⎝
1
words, f ( −1) = − , f ( 0 ) = −1 , f (1) = − e , and
e
(
)
⎧⎪ e x
f ( x) = ⎨
−x
⎪⎩e
if x < 0
if x ≥ 0
f ( 2 ) = −e 2 .
p⋅ 0
Therefore, f ( 0 ) = k ⋅ a ( )
−1 = k ⋅ a 0
−1 = k ⋅1
−1 = k
Domain: (−∞, ∞)
p⋅ 1
and f (1) = −a ( )
Range:
{ y | 0 < y ≤ 1}
or (0,1]
Intercept: (0,1)
−e = − a p
e = ap
Let’s use a = e, p = 1 . Then f ( x ) = −e x . Now
93.
we need to verify that this function yields the other
known points on the graph.
1
f ( −1) = −e−1 = −
e
x
⎪⎧ −e
f ( x) = ⎨
−x
⎪⎩−e
if x < 0
if x ≥ 0
f ( 2 ) = −e 2
So we have the function f ( x ) = −e x .
91.
−x
⎪⎧e
f ( x) = ⎨
x
⎪⎩ e
if x < 0
if x ≥ 0
Domain: (−∞, ∞)
Range:
{ y | −1 ≤ y < 0}
or [ −1, 0 )
Intercept: (0, −1)
94.
( −∞, ∞ )
Range: { y | y ≥ 1}
Intercept: ( 0,1)
−x
⎪⎧−e
f ( x) = ⎨
x
⎪⎩ −e
if x < 0
if x ≥ 0
Domain:
or [1, ∞ )
Domain:
( −∞, ∞ )
{ y | y ≤ −1}
Intercept: ( 0, −1)
Range:
or
( −∞, −1]
627
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
95. p(n) = 100(0.97) n
b.
a.
p (10) = 100(0.97)10 ≈ 74% of light
b.
p(25) = 100(0.97) 25 ≈ 47% of light
The probability that a car will arrive within 40
minutes of 12:00 PM is 0.982.
c.
96. p(h) = 760e −0.145 h
a.
F ( 40 ) = 1 − e −0.1(40) = 1 − e−4 ≈ 0.982
As t → ∞, F ( t ) = 1 − e−0.1t → 1 − 0 = 1
d. Graphing the function:
1
p(2) = 760e −0.145(2)
= 760e −0.290
≈ 568.68 mm of mercury
b.
p(10) = 760e −0.145(10)
0
= 760e −1.45
≈ 178.27 mm of mercury
e.
a.
p(3) = 16, 630(0.90)3 ≈ $12,123
b.
p(9) = 16, 630(0.90)9 ≈ $6, 443
A(3) = 100e −0.35(3)
a.
= 100e −3.5
≈ 3.02 square millimeters
b.
= 5e
F (15 ) = 1 − e −0.15(15) = 1 − e−2.25 ≈ 0.895
F ( 30 ) = 1 − e −0.15(30) = 1 − e−4.5 ≈ 0.989
The probability that a car will arrive within 30
minutes of 5:00 PM is 0.989.
99. D ( h ) = 5e −0.4 h
D (1) = 5e
40
The probability that a car will arrive within 15
minutes of 5:00 PM is 0.895.
A(10) = 100e −0.35(10)
−0.4
0
102. F ( t ) = 1 − e−0.15 t
= 100e −1.05
≈ 34.99 square millimeters
−0.4(1)
F ( 7 ) ≈ 0.50 , so about 7 minutes are needed
0
98. A(n) = A0 e−0.35 n
b.
40
for the probability to reach 50%.
1
97. p ( x) = 16, 630(0.90) x
a.
0
c.
≈ 3.35
As t → ∞, F ( t ) = 1 − e−0.15 t → 1 − 0 = 1
d. Graphing the function:
1
After 1 hours, 3.35 milligrams will be present.
D ( 6 ) = 5e −0.4(6) = 5e−2.4 ≈ 0.45 milligrams
After 6 hours, 0.45 milligrams will be present.
(
100. N = P 1 − e−0.15 d
)
0
(
)
= 1000 (1 − e
) ≈ 362
N ( 3) = 1000 1 − e −0.15(3)
e.
−0.45
0
40
F ( 6 ) ≈ 0.60 , so 6 minutes are needed for the
probability to reach 60%.
1
After 3 days, 362 students will have heard the
rumor.
101. F ( t ) = 1 − e−0.1t
a.
F (10 ) = 1 − e −0.1(10) = 1 − e −1 ≈ 0.632
The probability that a car will arrive within 10
minutes of 12:00 PM is 0.632.
0
0
40
628
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
20 x e −20
x!
103. P ( x) =
a.
b.
b.
2015 e −20
≈ 0.0516 or 5.16%
15!
The probability that 15 cars will arrive
between 5:00 PM and 6:00 PM is 5.16%.
2020 e −20
≈ 0.0888 or 8.88%
20!
The probability that 20 cars will arrive
between 5:00 PM and 6:00 PM is 8.88%.
⎛ 10 ⎞
− ⎜ ⎟1 ⎤
120 ⎡
⎢1 − e ⎝ 5 ⎠ ⎥ = 12 ⎡1 − e −2 ⎤ ≈ 10.376
⎣
⎦
10 ⎢
⎥
⎣
⎦
amperes after 1 second
⎛ 10 ⎞
I1 =
4 5 e −4
≈ 0.1563 or 15.63%
5!
The probability that 5 people will arrive within
the next minute is 15.63%.
b. As t → ∞, e
−
R
b.
R = 10⎝ 68+ 459.4
⎛
⎜
R
4221
−
4221
⎞
+2⎟
59 + 459.4 ⎠
4221
⎛ 4221
⎞
−
+2⎟
⎜
= 10⎝ T + 459.4 T + 459.4 ⎠
(
106. L ( t ) = 500 1 − e−0.0061 t
(
= 500 (1 − e
L ( 30 ) = 500 1 − e
⎛ 10 ⎞
≈ 70.95%
(
= 500 (1 − e
L ( 60 ) = 500 1 − e
See the graph at the top of the next page.
d.
I2 =
≈ 72.62%
−⎜ ⎟ 0.5 ⎤
120 ⎡
⎢1 − e ⎝ 10 ⎠ ⎥ = 24 ⎡1 − e −0.25 ⎤
⎣
⎦
5 ⎢
⎥
⎣
⎦
≈ 5.309 amperes after 0.5 second
⎛5⎞
= 102 = 100%
−⎜ ⎟1 ⎤
120 ⎡
⎢1 − e ⎝ 10 ⎠ ⎥ = 24 ⎡1 − e −0.5 ⎤
⎣
⎦
5 ⎢
⎥
⎣
⎦
≈ 9.443 amperes after 1 second
⎛5⎞
−0.0061(30)
)
−0.0061(60)
−0.366
⎛5⎞
−⎜ ⎟ 0.3 ⎤
120 ⎡
⎢1 − e ⎝ 10 ⎠ ⎥ = 24 ⎡1 − e −0.15 ⎤
⎣
⎦
5 ⎢
⎥
⎣
⎦
≈ 3.343 amperes after 0.3 second
I2 =
I2 =
)
≈ 84
The student will learn about 84 words after 30
minutes.
b.
c.
)
−0.183
→ 0 . Therefore, as,
t → ∞, I1 =
4221
⎞
+2⎟
D + 459.4 ⎠
4221
⎛ 4221
⎞
−
+2 ⎟
⎜
= 10⎝ 50 + 459.4 41+ 459.4 ⎠
⎛ 10 ⎞
−⎜ ⎟t
⎝5⎠
−⎜ ⎟t ⎤
120 ⎡
⎢1 − e ⎝ 5 ⎠ ⎥ → 12 [1 − 0] = 12 ,
10 ⎢
⎥
⎣
⎦
which means the maximum current is 12
amperes.
48 e −4
≈ 0.0298 or 2.98%
8!
The probability that 8 people will arrive within
the next minute is 2.98%.
P (8) =
a.
a.
−⎜ ⎟ 0.5 ⎤
120 ⎡
⎢1 − e ⎝ 5 ⎠ ⎥ = 12 ⎡1 − e −1 ⎤ ≈ 7.585
⎣
⎦
10 ⎢
⎥
⎣
⎦
amperes after 0.5 second
P ( 5) =
4221
I1 =
I1 =
4 x e −4
x!
⎛
⎜
−⎜ ⎟ 0.3 ⎤
120 ⎡
⎢1 − e ⎝ 5 ⎠ ⎥ = 12 ⎡1 − e −0.6 ⎤ ≈ 5.414
⎣
⎦
10 ⎢
⎥
⎣
⎦
amperes after 0.3 second
⎛ 10 ⎞
a.
P (20) =
105. R = 10⎝ T + 459.4
c.
⎛R⎞
P (15) =
104. P ( x) =
a.
−⎜ ⎟t ⎤
E⎡
107. I = ⎢1 − e ⎝ L ⎠ ⎥
R⎢
⎥
⎣
⎦
)
e.
As t → ∞, e
⎛5⎞
−⎜ ⎟t
⎝ 10 ⎠
→ 0 . Therefore, as,
−⎜ ⎟t ⎤
120 ⎡
⎢1 − e ⎝ 5 ⎠ ⎥ → 24 [1 − 0] = 24 ,
5 ⎢
⎥
⎣
⎦
which means the maximum current is 24
amperes.
⎛ 10 ⎞
)
t → ∞, I1 =
≈ 153
The student will learn about 153 words after
60 minutes.
f.
See the graph at the top of the next page.
629
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
⎛ −3000 ⎞
⎛ −t ⎞
E ⎜ ⎟
108. I = ⋅ e⎝ RC ⎠
R
⎛
a.
−0
120 ⎝⎜ 1000 ⋅2 ⎠⎟ 120 −1.5
I2 =
e
⋅e
=
≈ 0.0268
1000
1000
amperes after 3000 microseconds
⎞
120 ⎜⎝ 2000 ⋅1 ⎟⎠ 120 0
I1 =
e = 0.06
⋅e
=
2000
2000
amperes initially.
e.
The maximum current occurs at t = 0 .
Therefore, the maximum current is 0.12
amperes.
f.
Graphing the functions:
⎛ −1000 ⎞
120 ⎜⎝ 2000 ⋅1 ⎟⎠ 120 −1/ 2
e
⋅e
=
≈ 0.0364
2000
2000
amperes after 1000 microseconds
I1 =
⎛ −3000 ⎞
120 ⎜⎝ 2000 ⋅1 ⎟⎠ 120 −1.5
I1 =
e
⋅e
=
≈ 0.0134
2000
2000
amperes after 3000 microseconds
b. The maximum current occurs at t = 0 .
Therefore, the maximum current is 0.06
amperes.
c.
Graphing the function:
1 1 1
1
+ + + ... +
2! 3! 4!
n!
1 1 1
n = 4; 2 + + + ≈ 2.7083
2! 3! 4!
1 1 1 1 1
n = 6; 2 + + + + + ≈ 2.7181
2! 3! 4! 5! 6!
1 1 1 1 1 1 1
n = 8; 2 + + + + + + +
2! 3! 4! 5! 6! 7! 8!
≈ 2.7182788
1 1 1 1 1 1 1 1
1
n = 10; 2 + + + + + + + + +
2! 3! 4! 5! 6! 7! 8! 9! 10!
≈ 2.7182818
109. 2 +
⎛
d.
−0
⎞
120 ⎜⎝ 1000 ⋅2 ⎟⎠ 120 0
e = 0.12
⋅e
=
1000
1000
amperes initially.
I2 =
e ≈ 2.718281828
⎛ −1000 ⎞
120 ⎝⎜ 1000 ⋅2 ⎠⎟ 120 −1/ 2
⋅e
=
≈ 0.0728
I2 =
e
1000
1000
amperes after 1000 microseconds
630
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.1: Exponential Functions
110. 2 + 1 = 3
2+ 1
112.
f ( x) = a x
f ( A + B ) = a A+ B = a A ⋅ a B = f ( A) ⋅ f ( B )
= 2.5
1+1
2 +1
= 2.8
2+ 1
114.
f ( x) = a x ; f (α x) = aα x = ( a x ) = [ f ( x) ]
a.
3+3
2+ 1
(
)
(
1+ 1
2+ 2
3+ 3
4+4
≈ 2.717770035
b. Let Y1 =
1+ 1
α
)
(
≈ 2.721649485
2+ 1
=
1 x −x
e −e
2
f (− x) = sinh(− x)
1
= e− x − e x
2
1
= − e x − e− x
2
= − sinh x
= − f ( x)
Therefore, f ( x) = sinh x is an odd function.
115. sinh x =
2+ 2
a
x
α
= 2.7
1+ 1
1
f ( x)
f ( x) = a x ; f (− x) = a − x =
1+ 1
2+2
1
113.
2+ 2
)
(
)
1 x −x
e −e .
2
3.5
3+ 3
−6
4+ 4
6
5+5
2+ 1
≈ 2.718348855
−3.5
1+ 1
3+ 3
a.
4+ 4
5+ 5
6+6
e ≈ 2.718281828
111.
f ( x) = a x
f ( x + h) − f ( x ) a x + h − a x
=
h
h
a x ah − a x
=
h
x ( h
a a − 1)
=
h
⎛
ah − 1 ⎞
= ax ⎜
⎟
⎝ h ⎠
(
)
1 x
e + e− x
2
f (− x) = cosh(− x)
1
= e− x + e x
2
1 x
= e + e− x
2
= cosh x
= f ( x)
Thus, f ( x) = cosh x is an even function.
116. cosh x =
2+ 2
b. Let Y1 =
(
(
)
)
(
)
1 x
e + e− x .
2
6
−6
6
−1
631
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
c.
Section 7.2
(cosh x) 2 − (sinh x) 2
⎛ e x + e− x
= ⎜⎜
2
⎝
2
⎞ ⎛ e x − e− x
⎟⎟ − ⎜⎜
2
⎠ ⎝
⎞
⎟⎟
⎠
2
1. 3x − 7 ≤ 8 − 2 x
5 x ≤ 15
x≤3
The solution set is { x x ≤ 3} .
e2 x + 2 + e−2 x e2 x − 2 + e−2 x
−
4
4
e2 x + 2 + e−2 x − e2 x + 2 − e−2 x
=
4
4
=
4
=1
=
117.
( 2x ) + 1
f ( x) = 2
f (1) = 2
( 21 ) + 1 = 22 + 1 = 4 + 1 = 5
3.
1
4
4.
{x
[because (1/ 4) −1 = 4 ]
x > 0} or (0, ∞)
⎛1
⎞
5. ⎜ , −1⎟ , (1, 0 ) , ( a,1)
⎝a
⎠
( 22 ) + 1 = 24 + 1 = 16 + 1 = 17
f (2) = 2
6. 1
( 23 ) + 1 = 28 + 1 = 256 + 1 = 257
f (3) = 2
f (4) = 2
[because 23 = 8 ]
2. 3
7. False. If y = log a x , then x = a y .
( ) + 1 = 216 + 1 = 65,536 + 1 = 65,537
24
8. True
( 25 )
9. 9 = 32 is equivalent to 2 = log 3 9 .
f (5) = 2
+ 1 = 232 + 1 = 4, 294,967, 296 + 1
= 4, 294,967, 297
= 641× 6, 700, 417
10. 16 = 42 is equivalent to 2 = log 4 16 .
11. a 2 = 1.6 is equivalent to 2 = log a 1.6 .
118. Since the number of bacteria doubles every
minute, half of the container is full one minute
before it is full. Thus, it takes 59 minutes to fill
the container.
12. a3 = 2.1 is equivalent to 3 = log a 2.1 .
119 – 120. Answers will vary.
13. 2 x = 7.2 is equivalent to x = log 2 7.2 .
121. Given the function f ( x ) = a x , with a > 1 ,
14. 3x = 4.6 is equivalent to x = log3 4.6 .
If x > 0 , the graph becomes steeper as a
increases.
If x < 0 , the graph becomes less steep as a
increases.
15. e x = 8 is equivalent to x = ln 8 .
16. e2.2 = M is equivalent to 2.2 = ln M .
17. log 2 8 = 3 is equivalent to 23 = 8 .
122. Using the laws of exponents, we have:
a− x =
x
⎛1⎞
= ⎜ ⎟ . So y = a − x and
ax ⎝ a ⎠
1
1
⎛1⎞
18. log3 ⎜ ⎟ = − 2 is equivalent to 3−2 = .
9
⎝9⎠
x
⎛1⎞
y = ⎜ ⎟ will have the same graph.
⎝a⎠
19. log a 3 = 6 is equivalent to a 6 = 3 .
20. logb 4 = 2 is equivalent to b 2 = 4 .
21. log3 2 = x is equivalent to 3x = 2 .
632
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
22. log 2 6 = x is equivalent to 2 x = 6 .
39. F ( x) = log 2 x 2 requires x 2 > 0 .
x 2 > 0 for all x ≠ 0 .
The domain of F is { x x ≠ 0} .
x
23. ln 4 = x is equivalent to e = 4 .
24. ln x = 4 is equivalent to e4 = x .
40. H ( x) = log 5 x3 requires x3 > 0 .
x3 > 0 for all x > 0 .
The domain of H is { x x > 0} or ( 0, ∞ ) .
25. log 2 1 = 0 since 20 = 1 .
26. log8 8 = 1 since 81 = 8 .
41.
27. log5 25 = 2 since 52 = 25 .
1
⎛1⎞
28. log3 ⎜ ⎟ = − 2 since 3−2 = .
9
⎝9⎠
⎛1⎞
29. log1/ 2 16 = − 4 since ⎜ ⎟
⎝2⎠
⎛1⎞
30. log1/ 3 9 = − 2 since ⎜ ⎟
⎝ 3⎠
−4
= 24 = 16 .
2
=3 =9.
1
since 101/ 2 = 10 .
2
32. log5 3 25 =
2
since 52 / 3 = 251/ 3 = 3 25 .
3
33. log
2
4 = 4 since
( 2)
34. log
3
9 = 4 since
( 3)
42. g ( x ) = 8 + 5ln ( 2 x + 3) requires 2 x + 3 > 0 .
2x + 3 > 0
2 x > −3
3
x>−
2
⎧
3⎫
⎛ 3 ⎞
The domain of g is ⎨ x x > − ⎬ or ⎜ − , ∞ ⎟ .
2
⎝ 2 ⎠
⎩
⎭
−2
31. log10 10 =
43.
4
4
= 4.
= 9.
1
35. ln e = since e1/ 2 = e .
2
1
⎛ 1 ⎞
f ( x) = ln ⎜
>0.
⎟ requires
x +1
⎝ x +1⎠
1
p ( x) =
is undefined when x = −1 .
x +1
Interval
(−∞, −1)
(−1, ∞)
Test Value
−2
0
Value of p
−1
1
Conclusion negative positive
The domain of f is { x x > −1} or ( −1, ∞ ) .
36. ln e3 = 3 since e3 = e3 .
37.
x
⎡x ⎤
f ( x ) = 3 − 2 log 4 ⎢ − 5⎥ requires − 5 > 0 .
2
⎣2 ⎦
x
−5 > 0
2
x
>5
2
x > 10
The domain of f is { x x > 10} or (10, ∞ ) .
f ( x) = ln( x − 3) requires x − 3 > 0 .
x−3 > 0
x>3
1
⎛ 1 ⎞
>0.
44. g ( x) = ln ⎜
⎟ requires
x −5
⎝ x−5⎠
1
is undefined when x = 5 .
p ( x) =
x−5
The domain of f is { x x > 3} or ( 3, ∞ ) .
38. g ( x) = ln( x − 1) requires x − 1 > 0 .
x −1 > 0
x >1
Interval
Test Value
(−∞,5)
4
(5, ∞)
6
Value of p
−1
1
Conclusion negative positive
The domain of g is { x x > 1} or (1, ∞ ) .
The domain of g is { x x > 5} or ( 5, ∞ ) .
633
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
x +1
⎛ x +1⎞
45. g ( x) = log5 ⎜
>0.
⎟ requires
x
x
⎝
⎠
x +1
p ( x) =
is zero or undefined when
x
x = −1 or x = 0 .
Interval
(−∞, −1)
Test Value
−2
Value of p
Conclusion
1
2
positive
(−1, 0)
49. ln
50.
(0, ∞)
1
51.
−1
2
2
3 ≈ 4.055
52.
− 0.1
ln
negative positive
x
⎛ x ⎞
46. h( x) = log 3 ⎜
>0.
⎟ requires
−
x
1
x
−1
⎝
⎠
x
p ( x) =
is zero or undefined when
x −1
x = 0 or x = 1 .
Test Value
−1
(0,1)
−
1
2
2
1
−1
2
2
Conclusion positive negative positive
ln 4 + ln 2
≈ 2.303
log 4 + log 2
54.
log15 + log 20
≈ 0.434
ln15 + ln 20
55.
2 ln 5 + log 50
≈ −53.991
log 4 − ln 2
56.
3log 80 − ln 5
≈ 1.110
log 5 + ln 20
57. If the graph of f ( x) = log a x contains the point
(2, 2) , then f (2) = log a 2 = 2 . Thus,
log a 2 = 2
The domain of h is { x x < 0 or x > 1} ;
a2 = 2
( −∞, 0 ) ∪ (1, ∞ ) .
a=± 2
Since the base a must be positive by definition,
we have that a = 2 .
f ( x) = ln x requires ln x ≥ 0 and x > 0
ln x ≥ 0
58. If the graph of f ( x) = log a x contains the point
x ≥ e0
x ≥1
The domain of h is { x x ≥ 1} or [1, ∞ ) .
48. g ( x) =
53.
(1, ∞)
Value of p
47.
10
3 ≈ 30.099
0.04
1
2
−
( −∞, −1) ∪ ( 0, ∞ ) .
(−∞, 0)
ln 5
≈ 0.536
3
ln
The domain of g is { x x < −1 or x > 0} ;
Interval
5
≈ 0.511
3
⎛1
⎞
⎛1⎞
⎛1⎞
⎜ , −4 ⎟ , then f ⎜ ⎟ = log a ⎜ ⎟ = − 4 . Thus,
⎝2⎠
⎝2⎠
⎝2
⎠
1
⎛ ⎞
log a ⎜ ⎟ = − 4
⎝2⎠
1
a −4 =
2
1 1
=
a4 2
a4 = 2
1
requires ln x ≠ 0 and x > 0
ln x
ln x ≠ 0
x ≠ e0
x ≠1
The domain of h is { x x > 0 and x ≠ 1} ;
a = 21/ 4 ≈ 1.189
( 0,1) or (1, ∞ ) .
634
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
63. B
59.
64. F
65. D
66. H
67. A
68. C
69. E
70. G
71.
60.
f ( x) = ln( x + 4)
a.
Domain: (− 4, ∞)
b. Using the graph of y = ln x , shift the graph
4 units to the left.
61.
c.
d.
Range: (−∞, ∞)
Vertical Asymptote: x = − 4
f ( x) = ln( x + 4)
y = ln( x + 4)
x = ln( y + 4)
y+4=e
Inverse
x
y = ex − 4
f −1 ( x) = e x − 4
62.
e.
Range of f: (−∞, ∞)
f.
Shift the graph of y = e x down 4 units.
635
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
72.
f ( x) = ln( x − 3)
a.
73.
Domain: (3, ∞)
a.
b. Using the graph of y = ln x , shift the graph
3 units to the right.
c.
f ( x) = ln( x − 3)
y = ln( x − 3)
x = ln( y − 3)
y −3 = e
c.
d.
Inverse
x
y = ex + 3
f
−1
Domain: (0, ∞)
b. Using the graph of y = ln x, shift up 2 units.
Range: (−∞, ∞)
Vertical Asymptote: x = 3
d.
f ( x) = 2 + ln x
Range: (−∞, ∞)
Vertical Asymptote: x = 0
f ( x) = 2 + ln x
y = 2 + ln x
x = 2 + ln y
x − 2 = ln y
Inverse
y = e x−2
x
( x) = e + 3
f −1 ( x) = e x − 2
e.
Range of f: (−∞, ∞)
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = e x , shift the graph 3
units up.
f.
Using the graph of y = e x , shift the graph 2
units to the right.
636
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
74.
f ( x) = − ln(− x)
a.
75.
Domain: (− ∞, 0)
f ( x) = ln(2 x) − 3
a.
b. Using the graph of y = ln x , reflect the
graph about the y-axis, and reflect about the
x-axis.
Domain: (0, ∞)
b. Using the graph of y = ln x , compress the
graph horizontally by a factor of
1
, and
2
shift down 3 units.
c.
Range: (−∞, ∞)
Vertical Asymptote: x = 0
f ( x) = − ln(− x)
y = − ln(− x)
x = − ln(− y )
− x = ln(− y )
d.
c.
Inverse
d.
− y = e− x
y = −e
f
−1
−x
Range of f: (−∞, ∞)
f.
Using the graph of y = e x , reflect the graph
about the y-axis, and reflect about the x-axis.
f ( x) = ln(2 x) − 3
y = ln(2 x) − 3
x = ln(2 y ) − 3
x + 3 = ln(2 y )
Inverse
2 y = e x +3
1
y = e x +3
2
1
f −1 ( x) = e x + 3
2
( x ) = −e − x
e.
Range: (−∞, ∞)
Vertical Asymptote: x = 0
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = e x , reflect the graph
about the y-axis, and reflect about the x-axis.
637
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
76.
f ( x) = −2 ln( x + 1)
a.
77.
Domain: (−1, ∞)
a.
b. Using the graph of y = ln x , shift the graph
to the left 1 unit, reflect about the x-axis and
stretch vertically by a factor of 2.
c.
d.
Domain: (4, ∞)
b. Using the graph of y = log x , shift the graph
4 units to the right and 2 units up.
Range: (−∞, ∞)
Vertical Asymptote: x = −1
f ( x) = −2 ln( x + 1)
y = −2 ln( x + 1)
x = −2 ln( y + 1)
x
− = ln( y + 1)
2
y + 1 = e− x / 2
f ( x) = log ( x − 4 ) + 2
c.
Range: (−∞, ∞)
Vertical Asymptote: x = 4
d.
f ( x) = log ( x − 4 ) + 2
y = log ( x − 4 ) + 2
x = log ( y − 4 ) + 2
Inverse
Inverse
x − 2 = log ( y − 4 )
y − 4 = 10 x − 2
y = 10 x − 2 + 4
f −1 ( x) = 10 x − 2 + 4
y = e− x / 2 − 1
f −1 ( x) = e− x / 2 − 1
e.
Range of f: (−∞, ∞)
e.
Range of f: (−∞, ∞)
f.
f.
Using the graph of y = e x , reflect the graph
about the y-axis, stretch horizontally by a
factor of 2, and shift down 1 unit.
Using the graph of y = 10 x , shift the graph
2 units to the right and 4 units up.
638
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
78.
f ( x) =
a.
1
log x − 5
2
79.
a.
Domain: (0, ∞)
b. Using the graph of y = log x , compress the
b. Using the graph of y = log x , compress the
graph vertically by a factor of
1
, and
2
1
compress vertically by a factor of .
2
1
, and shift it
2
graph horizontally by a factor of
5 units down.
c.
d.
1
log ( 2 x )
2
Domain: (0, ∞)
f ( x) =
Range: (−∞, ∞)
Vertical Asymptote: x = 0
c.
1
log x − 5
2
1
y = log x − 5
2
1
x = log y − 5
2
1
x + 5 = log y
2
2( x + 5) = log y
d.
f ( x) =
Inverse
Range: (−∞, ∞)
Vertical Asymptote: x = 0
1
log ( 2 x )
2
1
y = log ( 2 x )
2
1
x = log ( 2 y )
2
2 x = log ( 2 y )
f ( x) =
Inverse
e.
Range of f: (−∞, ∞)
e.
2 y = 102 x
1
y = ⋅102 x
2
1
−1
f ( x ) = ⋅102 x
2
Range of f: (−∞, ∞)
f.
Using the graph of y = 10 x , shift the graph
5 units to the left, and compress horizontally
1
by a factor of .
2
f.
Using the graph of y = 10 x , compress the
y = 102( x + 5)
f −1 ( x) = 102( x + 5)
1
, and
2
1
compress vertically by a factor of .
2
graph horizontally by a factor of
639
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
80.
f ( x) = log(−2 x)
a.
81.
Domain: (−∞, 0)
a.
b. Using the graph of y = log x , reflect the
graph across the y-axis and compress
1
horizontally by a factor of .
2
c.
d.
Domain: (−2, ∞)
b. Using the graph of y = log3 x , shift 2 units
to the left, and shift up 3 units.
Range: (−∞, ∞)
Vertical Asymptote: x = 0
f ( x) = log(−2 x)
y = log(−2 x)
x = log(−2 y )
f ( x) = 3 + log 3 ( x + 2 )
c.
Range: (−∞, ∞)
Vertical Asymptote: x = −2
d.
f ( x ) = 3 + log 3 ( x + 2 )
y = 3 + log 3 ( x + 2 )
x = 3 + log 3 ( y + 2 )
Inverse
x − 3 = log3 ( y + 2 )
Inverse
y + 2 = 3x − 3
x
−2 y = 10
1
f −1 ( x) = − ⋅10 x
2
y = 3x − 3 − 2
f −1 ( x) = 3x −3 − 2
e.
Range of f: (−∞, ∞)
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = 10 x , reflect the
graph across the x-axis and compress
1
vertically by a factor of .
2
f.
Using the graph of y = 3x , shift 3 units to
the right, and shift down 2 units.
640
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
82.
f ( x) = 2 − log 3 ( x + 1)
a.
83.
Domain: (−1, ∞)
f ( x) = e x + 2 − 3
a.
b. Using the graph of y = log3 x , shift 1 unit to
the left, reflect the graph about the x-axis,
and shift 2 units up.
c.
Range: (−∞, ∞)
Vertical Asymptote: x = −1
d.
f ( x ) = 2 − log3 ( x + 1)
Domain: (−∞, ∞)
b. Using the graph of y = e x , shift the graph
two units to the left, and shift 3 units down.
c.
Range: (−3, ∞)
Horizontal Asymptote: y = −3
f ( x) = e x + 2 − 3
d.
y = ex+2 − 3
y = 2 − log3 ( x + 1)
x = 2 − log3 ( y + 1)
x = e y+2 − 3
Inverse
x+3= e
y + 2 = ln ( x + 3)
x − 2 = − log 3 ( y + 1)
2 − x = log3 ( y + 1)
y = ln ( x + 3) − 2
y + 1 = 32 − x
f
y = 32 − x − 1
f −1 ( x ) = 32 − x − 1
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = 3x , reflect the graph
about the y-axis, shift 2 units to the right,
and shift down 1 unit.
Inverse
y+2
−1
( x) = ln ( x + 3) − 2
e.
Range of f: (−3, ∞)
f.
Using the graph of y = ln x , shift 3 units to
the left, and shift down 2 units.
641
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
84.
f ( x) = 3e x + 2
a.
85.
Domain: (−∞, ∞)
a.
b. Using the graph of y = e x , stretch the graph
vertically by a factor of 3, and shift 2 units
up.
c.
d.
f ( x) = 2 x / 3 + 4
b. Using the graph of y = 2 x , stretch the graph
horizontally by a factor of 3, and shift 4
units up.
Range: (2, ∞)
Horizontal Asymptote: y = 2
c.
f ( x) = 3e x + 2
d.
x
y = 3e + 2
y
x = 3e + 2
x − 2 = 3e
x−2
= ey
3
Domain: (−∞, ∞)
Range: (4, ∞)
Horizontal Asymptote: y = 4
f ( x) = 2 x / 3 + 4
y = 2x / 3 + 4
Inverse
x = 2y/3 + 4
y
x−4= 2
y
= log 2 ( x − 4 )
3
y = 3log 2 ( x − 4 )
⎛ x−2⎞
y = ln ⎜
⎟
⎝ 3 ⎠
⎛ x−2⎞
f −1 ( x) = ln ⎜
⎟
⎝ 3 ⎠
e.
Range of f: (2, ∞)
f.
Using the graph of y = ln x , shift 3 units to
the left, and shift down 2 units.
Inverse
y/3
f −1 ( x ) = 3log 2 ( x − 4 )
e.
Range of f: (4, ∞)
f.
Using the graph of y = log 2 x , shift 4 units
to the right, and stretch vertically by a factor
of 3.
642
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
86.
87. log3 x = 2
f ( x) = −3x +1
a.
x = 32
x=9
The solution set is {9} .
Domain: (−∞, ∞)
b. Using the graph of y = 3x , shift the graph to
the left 1 unit, and reflect about the x-axis.
88. log5 x = 3
x = 53
x = 125
The solution set is {125} .
89. log 2 (2 x + 1) = 3
c.
2 x + 1 = 23
2x + 1 = 8
2x = 7
7
x=
2
Range: (−∞, 0)
Horizontal Asymptote: y = 0
⎧7 ⎫
The solution set is ⎨ ⎬ .
⎩2⎭
f ( x ) = −3x +1
d.
y = −3x +1
x = −3 y +1
90. log3 (3x − 2) = 2
Inverse
3 x − 2 = 32
3x − 2 = 9
3 x = 11
11
x=
3
− x = 3 y +1
y + 1 = log3 ( − x )
y = log3 ( − x ) − 1
f
−1
( x ) = log3 ( − x ) − 1
e.
Range of f: (−∞, 0)
f.
Using the graph of y = log3 x , reflect the
graph across the y-axis, and shift down 1
unit.
⎧11 ⎫
The solution set is ⎨ ⎬ .
⎩3⎭
91. log x 4 = 2
x2 = 4
x = 2 ( x ≠ − 2, base is positive)
The solution set is {2} .
⎛1⎞
92. log x ⎜ ⎟ = 3
⎝8⎠
1
x3 =
8
1
x=
2
⎧1 ⎫
The solution set is ⎨ ⎬ .
⎩2⎭
93. ln e x = 5
e x = e5
x=5
The solution set is {5} .
643
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
94. ln e−2 x = 8
−2 x
100. e −2 x =
8
e =e
− 2x = 8
x = −4
1
3
⎛1⎞
− 2 x = ln ⎜ ⎟
⎝3⎠
( )
−2 x = ln 3−1
The solution set is {−4} .
−2 x = − ln 3
2 x = ln 3
ln 3
x=
2
95. log 4 64 = x
4 x = 64
4 x = 43
x=3
The solution set is {3} .
⎧ ln 3 ⎫
⎬.
⎩ 2 ⎭
The solution set is ⎨
101.
96. log5 625 = x
5 x = 625
5 x = 54
x=4
The solution set is {4} .
e2 x +5 = 8
2 x + 5 = ln 8
2 x = −5 + ln 8
−5 + ln 8
x=
2
⎧ −5 + ln 8 ⎫
⎬.
2
⎩
⎭
The solution set is ⎨
97. log3 243 = 2 x + 1
32 x +1 = 243
2 x +1
102.
5
3
=3
2x + 1 = 5
2x = 4
x=2
The solution set is {2} .
e −2 x +1 = 13
−2 x + 1 = ln13
−2 x = −1 + ln13
−1 + ln13 1 − ln13
x=
=
−2
2
⎧1 − ln13 ⎫
⎬.
⎩ 2 ⎭
The solution set is ⎨
98. log 6 36 = 5 x + 3
(
)
65 x + 3 = 36
103. log3 x 2 + 1 = 2
65 x + 3 = 6 2
5x + 3 = 2
5 x = −1
1
x=−
5
x 2 + 1 = 32
x2 + 1 = 9
x2 = 8
x = ± 8 = ±2 2
{
}
The solution set is −2 2, 2 2 .
⎧ 1⎫
The solution set is ⎨− ⎬ .
⎩ 5⎭
(
)
104. log5 x 2 + x + 4 = 2
3x
99. e = 10
3 x = ln10
ln10
x=
3
2
x + x + 4 = 52
x 2 + x + 4 = 25
x 2 + x − 21 = 0
⎧ ln10 ⎫
The solution set is ⎨
⎬.
⎩ 3 ⎭
x=
−1 ±
12 − 4 (1)( −21)
2 (1)
=
−1 ± 85
2
⎧⎪ −1 − 85 −1 + 85 ⎫⎪
,
⎬.
2
2
⎪⎩
⎭⎪
The solution set is ⎨
644
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
109. 2 ⋅102 − x = 5
5
102 − x =
2
105. log 2 8 x = −3
8 x = 2 −3
(2 )
3 x
= 2 −3
3x
−3
2 − x = log
2 =2
3x = −3
x = −1
The solution set is {−1} .
x = 2 − log
⎧
⎩
5⎫
2⎭
⎧
⎩
5⎫
4⎭
110. 4e x +1 = 5
5
e x +1 =
4
5e0.2 x = 7
7
e0.2 x =
5
x + 1 = ln
5
4
The solution set is ⎨−1 + ln ⎬ .
111. a.
⎧
⎩
G ( x ) = log3 ( 2 x + 1)
We require that 2 x + 1 be positive.
2x + 1 > 0
2 x > −1
1
x>−
2
1⎫
⎛ 1 ⎞
⎧
Domain: ⎨ x | x > − ⎬ or ⎜ − , ∞ ⎟
2
⎝ 2 ⎠
⎩
⎭
7⎫
5⎭
The solution set is ⎨5ln ⎬ .
108. 8 ⋅102 x − 7 = 3
3
102 x − 7 =
8
3
8
2 x = 7 + log
5
4
x = −1 + ln
7
0.2 x = ln
5
⎛ 7⎞
5 ( 0.2 x ) = 5 ⎜ ln ⎟
⎝ 5⎠
7
x = 5ln
5
x=
5
2
The solution set is ⎨2 − log ⎬ .
3x = 3−1
x = −1
The solution set is {−1} .
2 x − 7 = log
5
2
− x = −2 + log
106. log3 3x = −1
107.
5
2
b.
G ( 40 ) = log3 ( 2 ⋅ 40 + 1)
= log3 81
=4
The point (40, 4) is on the graph of G.
3
8
1⎛
3⎞
⎜ 7 + log ⎟
2⎝
8⎠
c.
⎧1 ⎛
3 ⎞⎫
⎩ ⎝
⎠⎭
G ( x) = 2
log3 ( 2 x + 1) = 2
The solution set is ⎨ ⎜ 7 + log ⎟ ⎬ .
2
8
2 x + 1 = 32
2x +1 = 9
2x = 8
x=4
The point (4, 2) is on the graph of G.
645
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
112. a.
F ( x ) = log 2 ( x + 1) − 3
114.
We require that x + 1 be positive.
x +1 > 0
x > −1
Domain: { x | x > −1} or ( −1, ∞ )
b.
if x ≤ −1
⎧ ln(− x)
f ( x) = ⎨
−
−
−1 < x < 0
ln(
)
if
x
⎩
F ( 7 ) = log 2 ( 7 + 1) − 3
= log 2 ( 8 ) − 3
= 3−3
=0
The point ( 7, 0 ) is on the graph of F.
{ x | x < 0} ; ( −∞, 0 )
Range: { y | y ≥ 0} ; [ −, ∞ )
Intercept: ( −1, 0 )
Domain:
F ( x ) = −1
c.
log 2 ( x + 1) − 3 = −1
log 2 ( x + 1) = 2
x + 1 = 22
x +1 = 4
x=3
The point ( 3, −1) is on the graph of F.
113.
115.
⎧− ln x if 0 < x < 1
f ( x) = ⎨
if x ≥ 1
⎩ ln x
⎧ln ( − x ) if x < 0
f ( x) = ⎨
if x > 0
⎩ ln x
{ x | x > 0} ; ( 0, ∞ )
Range: { y | y ≥ 0} ; [ 0, ∞ )
Intercept: (1, 0 )
Domain:
{ x | x ≠ 0}
Range: ( −∞, ∞ )
Intercepts: ( −1, 0 ) , (1, 0 )
Domain:
116.
⎧ ln x if 0 < x < 1
f ( x) = ⎨
if x ≥ 1
⎩− ln x
{ x | x > 0} ; ( 0, ∞ )
Range: { y | y ≤ 0} ; ( −∞, 0]
Intercept: (1, 0 )
Domain:
646
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
117. pH = − log10 ⎡⎣ H + ⎤⎦
a.
pH = − log10 [ 0.1] = −(−1) = 1
b.
pH = − log10 [ 0.01] = −(−2) = 2
c.
pH = − log10 [ 0.001] = −(−3) = 3
d.
As the H + decreases, the pH increases.
e.
b.
3.5 = − log10 ⎡⎣ H + ⎤⎦
−3.5 = log10 ⎡⎣ H + ⎤⎦
120. A = A0 e −0.35 n
⎡ H + ⎤ = 10 −3.5
⎣ ⎦
≈ 3.16 × 10−4
= 0.000316
f.
a.
50 = 100e−0.35 n
0.5 = e −0.35n
ln ( 0.5 ) = −0.35n
7.4 = − log10 ⎡⎣ H + ⎤⎦
t=
ln ( 0.5 )
≈ 1.98
−0.35
Approximately 2 days.
−7.4 = log10 ⎡⎣ H + ⎤⎦
⎡ H + ⎤ = 10 − 7.4
⎣ ⎦
≈ 3.981× 10−8
= 0.00000003981
b.
10 = 100e−0.35 n
0.1 = e−0.35n
ln ( 0.1) = −0.35n
t=
ln ( 0.1)
≈ 6.58
−0.35
About 6.58 days, or 6 days and 14 hours.
118. H = − ( p1 log p1 + p2 log p2 + ⋅⋅⋅ + pn log pn )
= − p1 log p1 − p2 log p2 − ⋅⋅⋅ − pn log pn
a.
667 = 760e−0.145 h
667
= e−0.145h
760
⎛ 667 ⎞
ln ⎜
⎟ = −0.145h
⎝ 760 ⎠
⎛ 667 ⎞
ln ⎜
⎟
760 ⎠
≈ 0.90
h= ⎝
−0.145
Approximately 0.90 kilometers.
H = −0.014 log ( 0.014 ) − 0.041log ( 0.041)
121. F (t ) = 1 − e − 0.1t
− 0.124 log ( 0.124 ) − 0.128log ( 0.128 )
a.
− 0.003log ( 0.003) − 0.690 log ( 0.690 )
0.5 = 1 − e− 0.1t
− 0.5 = −e−0.1t
≈ 0.4283
b.
H max = log ( 6 ) ≈ 0.7782
0.5 = e−0.1t
ln ( 0.5 ) = − 0.1t
c.
E=
H
≈ 0.5504
H max
t=
≈ 6.93
− 0.1
Approximately 6.93 minutes.
119. p = 760e −0.145 h
a.
ln ( 0.5 )
b.
0.8 = 1 − e −0.1t
− 0.2 = −e−0.1t
320 = 760e−0.145 h
320
= e−0.145h
760
⎛ 320 ⎞
ln ⎜
⎟ = −0.145h
⎝ 760 ⎠
⎛ 320 ⎞
ln ⎜
⎟
760 ⎠
≈ 5.97
h= ⎝
−0.145
Approximately 5.97 kilometers.
0.2 = e −0.1t
ln ( 0.2 ) = − 0.1t
t=
ln ( 0.2 )
≈ 16.09
− 0.1
Approximately 16.09 minutes.
c.
It is impossible for the probability to reach
100% because e−0.1t will never equal zero;
thus, F (t ) = 1 − e− 0.1t will never equal 1.
647
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
122. F (t ) = 1 − e− 0.15 t
a.
0.50 = 1 − e
125. I =
− 0.15 t
Substituting E = 12 , R = 10 , L = 5 , and
I = 0.5 , we obtain:
12
− 10 / 5 t
0.5 = ⎡1 − e ( ) ⎤
⎦
10 ⎣
5
= 1 − e−2t
12
7
e −2 t =
12
−2t = ln ( 7 /12 )
− R/ L t
−0.5 = e ( )
0.5 = e−0.15t
ln ( 0.5 ) = −0.15t
t=
ln ( 0.5 )
≈ 4.62
−0.15
Approximately 4.62 minutes, or 4 minutes
and 37 seconds.
b.
0.80 = 1 − e− 0.15 t
−0.2 = −e−0.15t
t=
0.2 = e
ln ( 0.2 ) = −0.15t
ln ( 0.2 )
Substituting E = 12 , R = 10 , L = 5 , and
I = 1.0 , we obtain:
12
− 10 / 5 t
1.0 = ⎡1 − e ( ) ⎤
⎦
10 ⎣
10
= 1 − e −2t
12
1
e −2 t =
6
−2t = ln (1/ 6 )
≈ 10.73
−0.15
Approximately 10.73 minutes, or 10 minutes
and 44 seconds.
123.
D = 5e−0.4 h
2 = 5e−0.4 h
0.4 = e−0.4 h
ln ( 0.4 ) = − 0.4h
h=
ln ( 0.4 )
t=
(
N = P 1 − e−0.15 d
(
450 = 1000 1 − e
0.45 = 1 − e
)
−0.15 d
ln (1/ 6 )
≈ 0.8959
−2
It takes approximately 0.8959 second to
obtain a current of 0.5 ampere.
≈ 2.29
− 0.4
Approximately 2.29 hours, or 2 hours and 17
minutes.
124.
ln ( 7 /12 )
≈ 0.2695
−2
It takes approximately 0.2695 second to
obtain a current of 0.5 ampere.
−0.15t
t=
E⎡
− R/L t
1− e ( ) ⎤
⎦
R⎣
Graphing:
)
−0.15 d
− 0.55 = −e−0.15 d
0.55 = e−0.15 d
ln ( 0.55 ) = − 0.15 d
d=
ln ( 0.55 )
≈ 3.99
− 0.15
Approximately 4 days.
648
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.2: Logarithmic Functions
(
)
126. L(t ) = A 1 − e− k t
(
20 = 200 1 − e− k (5)
a.
0.1 = 1 − e
130. Intensity of car:
⎛ x ⎞
70 = 10 log ⎜ −12 ⎟
⎝ 10 ⎠
⎛ x ⎞
7 = log ⎜ −12 ⎟
⎝ 10 ⎠
x
107 = −12
10
x = 10−5
)
−5 k
−5 k
e
= 0.9
−5 k = ln 0.9
ln 0.9
k =−
≈ 0.0211
5
−( − ln 0.9 ) (10) ⎞
⎛
5
L(10) = 200 ⎜ 1 − e
⎟
⎝
⎠
b.
(
= 200 1 − e2ln 0.9
Intensity of truck is 10 ⋅10−5 = 10−4 .
⎛ 10−4 ⎞
L 10−4 = 10 log ⎜⎜ −12 ⎟⎟
⎝ 10 ⎠
(
)
( )
= 38 words
= 10 log 108
−( − ln 0.9 ) (15) ⎞
⎛
5
L(15) = 200 ⎜1 − e
⎟
⎝
⎠
c.
(
= 200 1 − e3ln 0.9
= 10 ⋅ 8
= 80 decibels
)
⎛ 125,892 ⎞
131. M (125,892) = log ⎜
⎟ ≈ 8.1
⎝ 10−3 ⎠
≈ 54 words
−( − ln 0.9 ) t ⎞
⎛
5
180 = 200 ⎜ 1 − e
⎟
⎝
⎠
d.
0.9 = 1 − e
e
ln 0.9 t
5
ln 0.9
t
5
(
127. L 10
−7
)
⎛ 7943 ⎞
132. M (7943) = log ⎜ −3 ⎟ ≈ 6.9
⎝ 10 ⎠
ln 0.9 t
5
133. R = ekx
a.
1.4 = e k (0.03)
= 0.1
= ln 0.1
t=
ln 0.1
ln 0.9
5
1.4 = e0.03 k
ln (1.4 ) = 0.03 k
≈ 109.27 minutes
⎛ 10−7
= 10 log ⎜⎜ −12
⎝ 10
( )
= 10 log 10
k=
⎞
⎟⎟
⎠
b.
5
c.
= 10 ⋅ 5
= 50 decibels
⎛ 10−1
128. L 10−1 = 10 log ⎜⎜ −12
⎝ 10
(
)
(
= 10 log 1011
)
⎛ 10−3
129. L 10−3 = 10 log ⎜⎜ −12
⎝ 10
)
( )
ln (1.4 )
0.03
≈ 11.216
R = e11.216(0.17) = e1.90672 ≈ 6.73
100 = e11.216 x
100 = e11.216 x
ln (100 ) = 11.216 x
⎞
⎟⎟
⎠
x=
= 10 ⋅ 9
= 90 decibels
11.216
≈ 0.41
5 = e11.216 x
ln 5 = 11.216 x
ln 5
≈ 0.143
x=
11.216
At a percent concentration of 0.143 or
higher, the driver should be charged with a
DUI.
e.
Answers will vary.
⎞
⎟⎟
⎠
= 10 log 109
ln (100 )
d.
= 10 ⋅11
= 110 decibels
(
)
649
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
134. No. Explanations will vary.
Age Depriciation rate
5
38, 000 = 21, 200e R ( 5)
38, 000
= e5 R
21, 200
135. If the base of a logarithmic function equals 1, we
would have the following:
f ( x ) = log 1 ( x )
f −1 ( x ) = 1x = 1 for every real number x.
In other words, f
−1
⎛ 38, 000 ⎞
ln ⎜
⎟ = 5R
⎝ 21, 200 ⎠
⎛ 38, 000 ⎞
ln ⎜
21, 200 ⎟⎠
R= ⎝
≈ 0.1167 ≈ 11.7%
5
would be a constant
function and, therefore, f −1 would not be oneto-one.
( )
136. New = Old e R t
Answers will vary.
Age Depriciation rate
R 1
1
38, 000 = 36, 600e ( )
38, 000
= eR
36, 600
Section 7.3
1. sum
⎛ 38, 000 ⎞
R = ln ⎜
⎟ ≈ 0.03754 ≈ 3.8%
⎝ 36, 600 ⎠
2
2. 7
R 2
38, 000 = 32, 400e ( )
38, 000
= e2 R
32, 400
⎛ 38, 000 ⎞
ln ⎜
⎟ = 2R
⎝ 32, 400 ⎠
3. r log a M
4. False
5. False
6. True
⎛ 38, 000 ⎞
ln ⎜
32, 400 ⎟⎠
R= ⎝
≈ 0.07971 ≈ 8%
2
3
7. log 3 371 = 71
R 3
38, 000 = 28, 750e ( )
38, 000
= e3 R
28, 750
⎛ 38, 000 ⎞
ln ⎜
⎟ = 3R
⎝ 28, 750 ⎠
8. log 2 2−13 = −13
9. ln e −4 = −4
10. ln e
⎛ 38, 000 ⎞
ln ⎜
28, 750 ⎟⎠
R= ⎝
≈ 0.0930 ≈ 9.3%
3
4
2
= 2
11. 2log2 7 = 7
12. eln 8 = 8
R 4
38, 000 = 25, 400e ( )
38, 000
= e4 R
25, 400
⎛ 38, 000 ⎞
ln ⎜
⎟ = 4R
⎝ 25, 400 ⎠
13. log8 2 + log8 4 = log8 ( 4 ⋅ 2 ) = log8 8 = 1
14. log 6 9 + log 6 4 = log 6 ( 9 ⋅ 4 )
= log 6 36
= log 6 62
⎛ 38, 000 ⎞
ln ⎜
24,500 ⎟⎠
R= ⎝
≈ 0.1007 ≈ 10.1%
4
=2
15. log 6 18 − log 6 3 = log 6
18
= log 6 6 = 1
3
650
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Properties of Logarithms
23. ln 6 = ln(2 ⋅ 3) = ln 2 + ln 3 = a + b
16
= log8 8 = 1
2
16. log8 16 − log8 2 = log8
24. ln
17. log 2 6 ⋅ log 6 4 = log 6 4log2 6
= log 6 ( 22 )
= log 6 2
log 2 6
2 log 2 6
= log 6 2log2 6
=2
26. ln 0.5 = ln
1
= ln1 − ln 2 = 0 − a = − a
2
= log8 ( 3
28. ln 27 = ln 33 = 3 ⋅ ln 3 = 3b
)
2 log3 8
29. ln 5 6 = ln 61/ 5
1
= ln 6
5
1
= ln ( 2 ⋅ 3)
5
1
= ( ln 2 + ln 3)
5
1
= (a + b)
5
= log8 3
2 log3 8
2
= log8 3log3 8
= log8 82
=2
5
4
3
= ln 3 − ln 2 = b − a
2
27. ln 8 = ln 23 = 3 ⋅ ln 2 = 3a
18. log 3 8 ⋅ log8 9 = log8 9log3 8
log3
25. ln1.5 = ln
2
= log 6 62
19. 3log3 5 − log3 4 = 3
2
= ln 2 − ln 3 = a − b
3
=
5
4
20. 5log5 6 + log5 7 = 5log5 (6⋅7) = 5log5 42 = 42
1/ 4
21. e
30. ln 4
log 2 16
e
Let a = log e2 16, then ( e 2 ) = 16.
a
e 2 a = 16
e2 a = 42
(e )
2 a 1/ 2
= ( 42 )
1/ 2
ea = 4
a = ln 4
Thus, e
22. e
log
e2
log 2 16
e
31. log 5 ( 25 x ) = log 5 25 + log 5 x = 2 + log 5 x
= eln 4 = 4 .
32. log 3
9
Let a = log e2 9, then ( e 2 ) = 9.
a
34. log 7 x5 = 5log 7 x
e 2 a = 32
2 a 1/ 2
x
x
= log 3 2 = log 3 x − log 3 32 = log 3 x − 2
9
3
33. log 2 z 3 = 3log 2 z
e2a = 9
(e )
2
⎛2⎞
= ln ⎜ ⎟
3
⎝3⎠
1 2
= ln
4 3
1
= ( ln 2 − ln 3)
4
1
= (a − b)
4
= ( 32 )
1/ 2
35. ln ( ex ) = ln e + ln x = 1 + ln x
e =3
a
a = ln 3
Thus, e
log
e2
9
=e
ln 3
36. ln
=3.
e
= ln e − ln x = 1 − ln x
x
37. ln ( xe x ) = ln x + ln e x = ln x + x
651
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
⎛ x
38. ln ⎜ x
⎝e
⎞
x
⎟ = ln x − ln e = ln x − x
⎠
1/ 3
47.
39. log a ( u 2 v 3 ) = log a u 2 + log a v 3
= 2 log a u + 3log a v
⎛ a
40. log 2 ⎜ 2
⎝b
(
1
= ⎡⎣ ln( x − 2)( x + 1) − ln( x + 4) 2 ⎤⎦
3
1
= [ ln( x − 2) + ln( x + 1) − 2 ln( x + 4) ]
3
1
1
2
= ln( x − 2) + ln( x + 1) − ln( x + 4)
3
3
3
⎞
2
⎟ = log 2 a − log 2 b = log 2 a − 2 log 2 b
⎠
)
41. ln x 2 1 − x = ln x 2 + ln 1 − x
= ln x 2 + ln(1 − x)1/ 2
2/3
⎡ ( x − 4 )2 ⎤
48. ln ⎢ 2
⎥
⎢⎣ x − 1 ⎥⎦
2
2 ⎡ ( x − 4) ⎤
= ln ⎢ 2
⎥
3 ⎣⎢ x − 1 ⎦⎥
1
= 2 ln x + ln(1 − x)
2
(
)
42. ln x 1 + x 2 = ln x + ln 1 + x 2
= ln x + ln (1 + x 2 )
1/ 2
2⎡
2
ln ( x − 4 ) − ln ( x 2 − 1) ⎤
⎦
3⎣
2
= ⎡⎣ 2 ln ( x − 4 ) − ln ( ( x + 1)( x − 1) ) ⎤⎦
3
2
= ⎡⎣ 2 ln ( x − 4 ) − ln ( x + 1) − ln ( x − 1) ⎤⎦
3
4
2
2
= ln ( x − 4 ) − ln ( x + 1) − ln ( x − 1)
3
3
3
=
1
= ln x + ln (1 + x 2 )
2
⎛ x3 ⎞
3
43. log 2 ⎜
⎟ = log 2 x − log 2 ( x − 3)
x
3
−
⎝
⎠
= 3log 2 x − log 2 ( x − 3)
⎛ 3 x2 + 1 ⎞
⎟
44. log 5 ⎜ 2
⎜ x −1 ⎟
⎝
⎠
= log 5 ( x 2 + 1)
1/ 3
⎡ x2 − x − 2 ⎤
ln ⎢
2 ⎥
⎣ ( x + 4) ⎦
1 ⎡ ( x − 2)( x + 1) ⎤
= ln ⎢
⎥
3 ⎣ ( x + 4) 2 ⎦
49. ln
− log 5 ( x 2 − 1)
5 x 1 + 3x
( x − 4)3
(
)
= ln 5 x 1 + 3 x − ln( x − 4)3
1
= log 5 ( x 2 + 1) − log 5 ( x 2 − 1)
3
1
= log 5 ( x 2 + 1) − log 5 ( ( x + 1)( x − 1) )
3
1
= log 5 ( x 2 + 1) − log 5 ( x + 1) − log 5 ( x − 1)
3
= ln 5 + ln x + ln 1 + 3 x − 3ln ( x − 4 )
= ln 5 + ln x + ln (1 + 3x )
1/ 2
1
= ln 5 + ln x + ln (1 + 3x ) − 3ln ( x − 4 )
2
⎡ 5x2 3 1 − x ⎤
50. ln ⎢
2 ⎥
⎣ 4( x + 1) ⎦
⎡ x( x + 2) ⎤
= log [ x( x + 2) ] − log( x + 3) 2
45. log ⎢
2 ⎥
⎣ ( x + 3) ⎦
= log x + log( x + 2) − 2 log( x + 3)
(
)
= ln 5 x 2 3 1 − x − ln ( 4( x + 1) 2 )
= ln 5 + ln x 2 + ln (1 − x )
1/ 3
⎡ x3 x + 1 ⎤
46. log ⎢
= log x3 x + 1 − log( x − 2) 2
2 ⎥
(
−
2)
x
⎣
⎦
= log x3 + log( x + 1)1/ 2 − 2 log( x − 2)
1
= 3log x + log( x + 1) − 2 log( x − 2)
2
(
− 3ln ( x − 4 )
)
2
− ⎡ ln 4 + ln ( x + 1) ⎤
⎣
⎦
1
= ln 5 + 2 ln x + ln (1 − x ) − ln 4 − 2 ln ( x + 1)
3
652
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Properties of Logarithms
⎛ x ⎞
⎛ x +1⎞
2
57. ln ⎜
⎟ + ln ⎜
⎟ − ln ( x − 1)
x
x
1
−
⎝
⎠
⎝
⎠
⎡ x x + 1⎤
2
= ln ⎢
⋅
⎥ − ln ( x − 1)
⎣ x −1 x ⎦
51. 3log 5 u + 4 log 5 v = log 5 u 3 + log 5 v 4
= log 5 ( u v
3 4
)
52. 2 log 3 u − log 3 v = log 3 u 2 − log 3 v
⎡ x +1
⎤
= ln ⎢
÷ ( x 2 − 1) ⎥
⎣ x −1
⎦
⎡
⎤
x +1
⎥
= ln ⎢
⎢⎣ ( x − 1) ( x 2 − 1) ⎥⎦
⎡
⎤
x +1
= ln ⎢
⎥
−
−
+
(
1)(
1)(
1)
x
x
x
⎣
⎦
⎛ u2 ⎞
= log 3 ⎜ ⎟
⎝ v ⎠
⎛ x⎞
53. log 3 x − log 3 x3 = log 3 ⎜⎜ 3 ⎟⎟
⎝ x ⎠
⎛ x1/ 2 ⎞
= log 3 ⎜ 3 ⎟
⎝ x ⎠
⎛ 1 ⎞
= ln ⎜
2 ⎟
⎝ ( x − 1) ⎠
= ln( x − 1) −2
= − 2 ln( x − 1)
= log 3 x −5 / 2
=−
⎛1⎞
⎛ 1
54. log 2 ⎜ ⎟ + log 2 ⎜ 2
⎝ x⎠
⎝x
5
log 3 x
2
⎞
⎟ = log 2
⎠
⎛1 1 ⎞
⎜ ⋅ 2⎟
⎝x x ⎠
⎛ 1 ⎞
= log 2 ⎜ 3 ⎟
⎝x ⎠
= log 2 x −3
⎛ x2 + 2 x − 3 ⎞
⎛ x2 + 7 x + 6 ⎞
58. log ⎜
−
log
⎟
⎜
⎟
2
⎝ x −4 ⎠
⎝ x+2 ⎠
⎡ ⎛ x2 + 2x − 3 ⎞ ⎤
⎢⎜
⎟⎥
x2 − 4 ⎠ ⎥
= log ⎢ ⎝ 2
⎢⎛ x + 7x + 6 ⎞ ⎥
⎢⎜
⎟⎥
⎣⎢ ⎝ x + 2 ⎠ ⎦⎥
= −3log 2 x
55. log 4 ( x 2 − 1) − 5log 4 ( x + 1)
= log 4 ( x − 1) − log 4 ( x + 1)
2
⎡ ( x + 3)( x − 1)
⎤
x+2
= log ⎢
⋅
⎥
⎣ ( x − 2)( x + 2) ( x + 6)( x + 1) ⎦
⎡ ( x + 3)( x − 1) ⎤
= log ⎢
⎥
⎣ ( x − 2)( x + 6)( x + 1) ⎦
5
⎡ x2 −1 ⎤
= log 4 ⎢
⎥
5
⎢⎣ ( x + 1) ⎥⎦
⎡ ( x + 1)( x − 1) ⎤
= log 4 ⎢
⎥
5
⎢⎣ ( x + 1)
⎥⎦
⎡ x −1 ⎤
= log 4 ⎢
⎥
4
⎢⎣ ( x + 1) ⎥⎦
⎛4⎞
59. 8log 2 3x − 2 − log 2 ⎜ ⎟ + log 2 4
⎝ x⎠
= log 2
3x − 2
)
8
− ( log 2 4 − log 2 x ) + log 2 4
= log 2 (3 x − 2) 4 − log 2 4 + log 2 x + log 2 4
= log 2 (3 x − 2) 4 + log 2 x
56. log ( x 2 + 3x + 2 ) − 2 log 2 ( x + 1)
= log ( x 2 + 3 x + 2 ) − log 2 ( x + 1)
(
= log 2 ⎡⎣ x(3 x − 2) 4 ⎤⎦
2
60. 21log 3 3 x + log 3 ( 9 x 2 ) − log 3 9
⎛ x 2 + 3x + 2 ⎞
⎟
= log ⎜
2
⎜
⎟
⎝ ( x + 1) ⎠
⎛ ( x + 2 )( x + 1) ⎞
⎟
= log ⎜
⎜ ( x + 1)2 ⎟
⎝
⎠
⎛ x+2⎞
= log ⎜
⎟
⎝ x +1 ⎠
= log 3 ( x1/ 3 ) + log 3 ( 9 ) + log 3 ( x 2 ) − log 3 9
21
= log 3 x 7 + log 3 x 2
= log 3 ( x 7 ⋅ x 2 )
= log 3 ( x 9 )
653
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
1
61. 2 log a ( 5 x 3 ) − log a (2 x + 3)
2
70. log 5 8 =
= log a ( 5 x 3 ) − log a (2 x + 3)1/ 2
log 8
log 5
≈ 2.584
2
62.
ln e
≈ 0.874
ln π
= log a ( 25 x 6 ) − log a 2 x + 3
71. log π e =
⎡ 25 x 6 ⎤
= log a ⎢
⎥
⎣ 2x + 3 ⎦
72. log π 2 =
1
1
log ( x 3 + 1) + log ( x 2 + 1)
3
2
73. y = log 4 x =
= log ( x3 + 1)
1/ 3
+ log ( x 2 + 1)
1/ 2
3
ln x
log x
or y =
ln 4
log 4
2
= log ⎡ x + 1 ⋅ x + 1 ⎤
⎣
⎦
3
ln 2
≈ 0.303
ln π
2
4
−1
63. 2 log 2 ( x + 1) − log 2 ( x + 3) − log 2 ( x − 1)
= log 2 ( x + 1) − log 2 ( x + 3) − log 2 ( x − 1)
2
−2
( x + 1)
− log 2 ( x − 1)
( x + 3)
⎡ ( x + 1)2 ⎤
= log 2 ⎢
⎥
⎢⎣ ( x + 3)( x − 1) ⎥⎦
2
= log 2
74. y = log 5 x =
ln x
log x
or y =
ln 5
log 5
2
0
3
64. 3log 5 ( 3x + 1) − 2 log 5 ( 2 x − 1) − log 5 x
−2
= log 5 ( 3 x + 1) − log 5 ( 2 x − 1) − log 5 x
3
2
( 3x + 1)
= log 5
− log 5 x
2
( 2 x − 1)
⎡ ( 3x + 1)3 ⎤
= log 5 ⎢
⎥
2
⎢⎣ x ( 2 x − 1) ⎥⎦
3
65. log 3 21 =
log 21
≈ 2.771
log 3
66. log 5 18 =
log18
≈ 1.796
log 5
75. y = log 2 ( x + 2) =
3
−2
76. y = log 4 ( x − 3) =
0
68. log1/ 2 15 =
log15
log15
=
≈ −3.907
log (1/ 2 ) − log 2
−2
7=
log 7
log 2
ln( x − 3)
log( x − 3)
or y =
ln 4
log 4
2
log 71
log 71
=
≈ −3.880
log (1/ 3) − log 3
2
5
−2
67. log1/ 3 71 =
69. log
ln( x + 2)
log( x + 2)
or y =
ln 2
log 2
8
≈ 5.615
654
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Properties of Logarithms
77. y = log x −1 ( x + 1) =
ln( x + 1)
log( x + 1)
or y =
ln( x − 1)
log( x − 1)
d.
( f D h )( x ) = f ( h ( x ) ) = log 2 ( 4 x )
or
= log 2 4 + log 2 x = 2 + log 2 x
4
Domain:
5
0
e.
{ x | x > 0}
or
( 0, ∞ )
( f D h )(8) = log 2 ( 4 ⋅ 8) = log 2 32 = 5
or
= 2 + log 2 8 = 2 + 3 = 5
−4
78. y = log x + 2 ( x − 2) =
ln( x − 2)
log( x − 2)
or y =
ln( x + 2)
log( x + 2)
81. ln y = ln x + ln C
ln y = ln ( xC )
2
y = Cx
8
0
82. ln y = ln ( x + C )
y = x+C
−2
79.
83. ln y = ln x + ln( x + 1) + ln C
f ( x ) = ln x ; g ( x ) = e x ; h ( x ) = x 2
a.
b.
ln y = ln ( x ( x + 1) C )
( f D g )( x ) = f ( g ( x ) ) = ln ( e x ) = x
Domain: { x | x is any real number} or
( −∞, ∞ )
y = Cx ( x + 1)
84. ln y = 2 ln x − ln ( x + 1) + ln C
⎛ x 2C ⎞
ln y = ln ⎜
⎟
⎝ x +1⎠
Cx 2
y=
x +1
( g D f )( x ) = g ( f ( x ) ) = eln x = x
Domain: { x | x > 0} or ( 0, ∞ )
(Note: the restriction on the domain is due to
the domain of ln x )
c.
d.
e.
80.
( f D g )( 5 ) = 5
85. ln y = 3x + ln C
[from part (a)]
( f D h )( x ) = f ( h ( x ) ) = ln ( x 2 )
Domain: { x | x ≠ 0}
( f D h )( e ) = ln ( e
2
ln y = ln e3 x + ln C
ln y = ln ( Ce3 x )
y = Ce3 x
) = 2 ln e = 2 ⋅1 = 2
86. ln y = − 2 x + ln C
f ( x ) = log 2 x ; g ( x ) = 2 ; h ( x ) = 4 x
x
a.
b.
ln y = ln e −2 x + ln C
ln y = ln ( Ce −2 x )
( f D g )( x ) = f ( g ( x ) ) = log 2 ( 2 x ) = x
Domain: { x | x is any real number} or
( −∞, ∞ )
y = Ce −2 x
87. ln ( y − 3) = − 4 x + ln C
=x
( g D f )( x ) = g ( f ( x ) ) = 2
Domain: { x | x > 0} or ( 0, ∞ )
ln ( y − 3) = ln e −4 x + ln C
log 2 x
ln ( y − 3) = ln ( Ce−4 x )
(Note: the restriction on the domain is due to
the domain of log 2 x )
c.
( f D g )( 3) = 3
y − 3 = Ce−4 x
y = Ce−4 x + 3
[from part (a)]
655
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
88. ln ( y + 4 ) = 5 x + ln C
log 4 log 6 log 8
⋅
⋅
log 2 log 4 log 6
log 8
=
log 2
92. log 2 4 ⋅ log 4 6 ⋅ log 6 8 =
ln ( y + 4 ) = ln e + ln C
5x
ln ( y + 4 ) = ln ( Ce5 x )
y + 4 = Ce5 x
log 23
log 2
3log 2
=
log 2
=3
=
y = Ce − 4
5x
89. 3ln y =
1
1
ln ( 2 x + 1) − ln ( x + 4 ) + ln C
2
3
ln y 3 = ln ( 2 x + 1)
− ln ( x + 4 )
1/ 2
1/ 3
+ ln C
93. log 2 3 ⋅ log 3 4 ⋅ ⋅⋅ log n ( n + 1) ⋅ log n +1 2
⎡ C ( 2 x + 1)1/ 2 ⎤
ln y 3 = ln ⎢
⎥
1/ 3
⎢⎣ ( x + 4 )
⎥⎦
=
C ( 2 x + 1)
1/ 2
y3 =
log 2
log 2
=1
=
( x + 4)
1/ 3
⎡ C ( 2 x + 1)1/ 2 ⎤
y=⎢
⎥
1/ 3
⎢⎣ ( x + 4 )
⎥⎦
3
y=
1/ 3
94. log 2 2 ⋅ log 2 4 ⋅ . . . ⋅ log 2 2n
C ( 2 x + 1)
1/ 6
= log 2 2 ⋅ log 2 22 ⋅⋅ ⋅ log 2 2n
( x + 4)
1/ 9
= 1 ⋅ 2 ⋅ 3 ⋅⋅ ⋅ n
= n!
1
1
90. 2 ln y = − ln x + ln ( x 2 + 1) + ln C
2
3
ln y = − ln x
2
1/ 2
+ ln ( x + 1)
1/ 3
2
⎡ C ( x 2 + 1)
ln y 2 = ln ⎢
⎢
x1/ 2
⎣
1/ 3
(
y =
+ ln C
(
)
(
)(
)
= log a ⎡ x + x 2 − 1 x − x 2 − 1 ⎤
⎣⎢
⎦⎥
2
2
= log a ⎡⎣ x − ( x − 1) ⎤⎦
= log a ⎡⎣ x 2 − x 2 + 1⎤⎦
⎤
⎥
⎥
⎦
C ( x 2 + 1)
= log a 1
x1/ 2
⎡ C ( x 2 + 1)1/ 3 ⎤
⎢
⎥
y=
⎢
⎥
x1/ 2
⎣
⎦
=0
1/ 2
96. log a
(
)
x + x − 1 + log a
(
= log a ⎡ x + x − 1
⎣
= log a ⎡⎣ x − ( x − 1) ⎤⎦
C ( x 2 + 1)
1/ 6
y=
)
95. log a x + x 2 − 1 + log a x − x 2 − 1 :
1/ 3
2
log 3 log 4 log ( n + 1)
log 2
⋅
⋅⋅ ⋅
⋅
log 2 log 3
log n
log ( n + 1)
x1/ 4
)(
(
x − x −1
)
)
x − x −1 ⎤
⎦
= log a [ x − x + 1]
91. log 2 3 ⋅ log 3 4 ⋅ log 4 5 ⋅ log 5 6 ⋅ log 6 7 ⋅ log 7 8
= log a 1
log 3 log 4 log 5 log 6 log 7 log 8
=
⋅
⋅
⋅
⋅
⋅
log 2 log 3 log 4 log 5 log 6 log 7
=0
log 8 log 23
=
log 2 log 2
3log 2
=
log 2
=3
=
656
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.3: Properties of Logarithms
97. 2 x + ln (1 + e −2 x ) = ln e 2 x + ln (1 + e−2 x )
(
= ln e2 x (1 + e −2 x )
= ln ( e + e
2x
0
= ln ( e2 x + 1)
98.
)
103. If A = log a M and B = log a N , then a A = M
)
and a B = N .
⎛M
log a ⎜
⎝N
= A− B
= log a M − log a N
f ( x + h) − f ( x) log a ( x + h) − log a x
=
h
h
⎛ x+h⎞
log a ⎜
⎟
⎝ x ⎠
=
h
1
⎛ h⎞
= ⋅ log a ⎜ 1 + ⎟
h
⎝ x⎠
⎛1⎞
104. log a ⎜ ⎟ = log a N −1
⎝N⎠
= −1 ⋅ log a N
= − log a N ,
1
5 −5
5
y = log x 2
y = 2 log x
−3
−3
The domain of Y1 = log a x is { x x ≠ 0} . The
2
domain of Y2 = 2 log a x is { x x > 0} . These two
domains are different because the logarithm
property log a x n = n ⋅ log a x holds only when
log a x exists.
− log1/ a x = f ( x)
− f ( x) = log1/ a x
106. Answers may vary. One possibility follows:
Let a = 2 , x = 8 , and r = 3 . Then
f ( AB ) = log a ( AB )
= log a A + log a B
( log a x )
= f ( A) + f ( B )
r
= ( log 2 8 ) = 33 = 27 . But
3
r log a x = 3log 2 8 = 3 ⋅ 3 = 9 . Thus,
( log 2 8) ≠ 3log 2 8 and, in general,
r
( log a x ) ≠ r log a x .
f ( x) = log a x
3
⎛1⎞
⎛1⎞
f ⎜ ⎟ = log a ⎜ ⎟
⎝x⎠
⎝ x⎠
= log a 1 − log a x
107. Answers may vary. One possibility follows:
Let x = 4 and y = 4 . Then
= − log a x
log 2 ( x + y ) = log 2 (4 + 4) = log 2 8 = 3 . But
log 2 x + log 2 y = log 2 4 + log 2 4 = 2 + 2 = 4 .
Thus, log 2 (4 + 4) ≠ log 2 4 + log 2 4 and, in
general, log 2 ( x + y ) ≠ log 2 x + log 2 y .
= − f ( x)
102.
2
−5
f ( x) = log a x means that x = a f ( x ) .
Now, raising both sides to the −1 power, we
f ( x)
f ( x)
−1
⎛1⎞
obtain x −1 = ( a f ( x ) ) = ( a −1 )
=⎜ ⎟ .
⎝a⎠
f ( x)
101.
Y2 = 2 log x
2
⎛1⎞
means that log1/ a x −1 = f ( x) .
x −1 = ⎜ ⎟
⎝a⎠
Thus, log1/ a x −1 = f ( x)
100.
a ≠1
105. Y1 = log x 2
⎛ h ⎞h
= log a ⎜1 + ⎟ , h ≠ 0
⎝ x⎠
99.
⎛ aA ⎞
⎞
=
log
a ⎜ B ⎟
⎟
⎠
⎝a ⎠
= log a a A − B
f ( x ) = log a x
f ( xα ) = log a xα = α log a x = α f ( x)
108. No. log 3 (−5) does not exist.
657
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
5. log 4 x = 2
Section 7.4
1.
x = 42
x = 16
The solution set is {16} .
2
x − 7 x − 30 = 0
( x + 3)( x − 10) = 0
x + 3 = 0 or x − 10 = 0
x = −3 or
x = 10
The solution set is {−3, 10} .
6. log ( x + 6) = 1
x + 6 = 101
x + 6 = 10
x=4
The solution set is {4} .
2. Let u = x + 3 . Then
( x + 3) 2 − 4( x + 3) + 3 = 0
u 2 − 4u + 3 = 0
(u − 1)(u − 3) = 0
7. log 2 (5 x) = 4
u − 1 = 0 or u − 3 = 0
u = 1 or
u=3
Back substituting u = x + 3 , we obtain
x +3 =1
or x + 3 = 3
x = −2 or
x=0
The solution set is {−2, 0} .
5 x = 24
5 x = 16
16
x=
5
⎧16 ⎫
⎩5⎭
The solution set is ⎨ ⎬ .
8. log3 (3x − 1) = 2
3. x3 = x 2 − 5
Using INTERSECT to solve:
y1 = x3 ; y2 = x 2 − 5
3 x − 1 = 32
3x − 1 = 9
3 x = 10
10
x=
3
10
−5
5
⎧10 ⎫
⎩3⎭
The solution set is ⎨ ⎬ .
9. log 4 ( x + 2) = log 4 8
x+2=8
x=6
The solution set is {6} .
−10
Thus, x ≈ −1.43 , so the solution set is {−1.43} .
4. x3 − 2 x + 2 = 0
Using ZERO to solve: y1 = x3 − 2 x + 2
10. log5 (2 x + 3) = log 5 3
2x + 3 = 3
2x = 0
x=0
The solution set is {0} .
5
−5
5
−5
Thus, x ≈ −1.77 , so the solution set is {−1.77} .
11.
1
log 3 x = 2 log3 2
2
log3 x1/ 2 = log3 22
x1/ 2 = 4
x = 16
The solution set is {16} .
658
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
12. − 2 log 4 x = log 4 9
log 4 x
−2
16. 2 log3 ( x + 4) − log 3 9 = 2
= log 4 9
log3 ( x + 4) 2 − log3 32 = 2
x −2 = 9
1
=9
x2
1
x2 =
9
log 3 ( x + 4) 2 − 2 = 2
log 3 ( x + 4)2 = 4
( x + 4)2 = 34
( x + 4)2 = 81
x + 4 = ±9
x = −4±9
x = 5 or x = −13
Since log3 ( −13 + 4 ) = log3 ( − 9 ) is undefined,
1
3
⎛ 1⎞
Since log 4 ⎜ − ⎟ is undefined, the solution set is
⎝ 3⎠
1
⎧ ⎫
⎨ ⎬.
⎩3⎭
x=±
the solution set is {5} .
17. log x + log( x + 15) = 2
log ( x( x + 15) ) = 2
13. 3log 2 x = − log 2 27
log 2 x3 = log 2 27 −1
x( x + 15) = 102
−1
3
x = 27
1
x3 =
27
1
x=
3
x 2 + 15 x − 100 = 0
( x + 20)( x − 5) = 0
x = − 20 or x = 5
Since log ( −20 ) is undefined, the solution set is
{5} .
⎧1 ⎫
The solution set is ⎨ ⎬ .
⎩3⎭
18. log x + log( x − 21) = 2
log ( x( x − 21) ) = 2
14. 2 log 5 x = 3log5 4
x( x − 21) = 102
log5 x 2 = log5 43
x 2 − 21x − 100 = 0
( x + 4)( x − 25) = 0
x = −4 or x = 25
Since log ( −4 ) is undefined, the solution set is
x 2 = 64
x = ±8
Since log 5 ( − 8 ) is undefined, the solution set is
{8} .
{25} .
15. 3log 2 ( x − 1) + log 2 4 = 5
19.
log 2 ( x − 1)3 + log 2 4 = 5
(
)
log 2 4( x − 1)3 = 5
log(2 x + 1) = 1 + log( x − 2)
log(2 x + 1) − log( x − 2) = 1
⎛ 2x +1 ⎞
log ⎜
⎟ =1
⎝ x−2 ⎠
2x +1
= 101
x−2
2 x + 1 = 10( x − 2)
2 x + 1 = 10 x − 20
−8 x = −21
−21 21
x=
=
−8
8
3
4( x − 1) = 25
32
( x − 1)3 =
4
( x − 1)3 = 8
x −1 = 2
x=3
The solution set is {3} .
⎧ 21 ⎫
⎬.
⎩8⎭
The solution set is ⎨
659
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
20. log(2 x) − log( x − 3) = 1
24.
⎛ 2x ⎞
log ⎜
⎟ =1
⎝ x −3⎠
2x
= 101
x −3
2 x = 10( x − 3)
2 x = 10 x − 30
−8 x = −30
−30 15
x=
=
4
−8
log 5 ( x + 3) = 1 − log5 ( x − 1)
log5 ( x + 3) + log5 ( x − 1) = 1
log 5 [ ( x + 3)( x − 1)] = 1
( x + 3)( x − 1) = 51
x 2 − x + 3x − 3 = 5
x2 + 2 x − 8 = 0
( x + 4)( x − 2) = 0
x = −4 or x = 2
Since log 5 ( − 4 + 3) = log 5 ( − 1) is undefined, the
solution set is {2}.
⎧15 ⎫
⎩4⎭
The solution set is ⎨ ⎬ .
25. ln x + ln( x + 2) = 4
ln ( x( x + 2) ) = 4
21. log 2 ( x + 7) + log 2 ( x + 8) = 1
log 2 [ ( x + 7)( x + 8) ] = 1
x( x + 2) = e 4
x 2 + 2 x − e4 = 0
( x + 7)( x + 8) = 21
x 2 + 8 x + 7 x + 56 = 2
x=
x 2 + 15 x + 54 = 0
( x + 9)( x + 6) = 0
x = −9 or x = −6
Since log 2 ( − 9 + 7 ) = log 2 ( − 2 ) is undefined,
the solution set is {−6} .
2(1)
=
− 2 ± 4 + 4e 4
2
=
− 2 ± 2 1 + e4
2
= −1 ± 1 + e 4
22. log 6 ( x + 4) + log 6 ( x + 3) = 1
x = −1 − 1 + e 4 or x = −1 + 1 + e 4
≈ −8.456
≈ 6.456
Since ln ( −8.456 ) is undefined, the solution set
log 6 [ ( x + 4)( x + 3) ] = 1
( x + 4)( x + 3) = 61
x 2 + 3 x + 4 x + 12 = 6
{
}
is −1 + 1 + e 4 ≈ {6.456} .
x2 + 7 x + 6 = 0
( x + 6)( x + 1) = 0
x = −6 or x = −1
Since log 6 ( − 6 + 4 ) = log 6 ( − 2 ) is undefined,
26. ln( x + 1) − ln x = 2
⎛ x +1⎞
ln ⎜
⎟=2
⎝ x ⎠
x +1 2
=e
x
x + 1 = e2 x
the solution set is {−1} .
23.
− 2 ± 22 − 4(1)(− e 4 )
log8 ( x + 6) = 1 − log8 ( x + 4)
log8 ( x + 6) + log8 ( x + 4) = 1
log8 [ ( x + 6)( x + 4) ] = 1
e2 x − x = 1
(
)
x e2 − 1 = 1
( x + 6)( x + 4) = 81
2
x + 4 x + 6 x + 24 = 8
x=
1
≈ 0.157
e −1
⎧ 1 ⎫
The solution set is ⎨ 2 ⎬ ≈ {0.157} .
⎩ e − 1⎭
2
x + 10 x + 16 = 0
( x + 8)( x + 2) = 0
x = −8 or x = −2
Since log8 ( − 8 + 6 ) = log8 ( − 2 ) is undefined, the
2
solution set is {−2} .
660
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
27. log3 ( x + 1) + log 3 ( x + 4 ) = 2
29. log1/ 3 ( x 2 + x) − log1/ 3 ( x 2 − x) = −1
log 3 ⎡⎣( x + 1)( x + 4 ) ⎤⎦ = 2
⎛ x2 + x ⎞
log1/ 3 ⎜⎜ 2
⎟⎟ = −1
⎝x −x⎠
( x + 1)( x + 4 ) = 32
x2 + 4 x + x + 4 = 9
x2 + x
⎛1⎞
=⎜ ⎟
x2 − x ⎝ 3 ⎠
x2 + 5x − 5 = 0
x=
x2 + x
− 5 ± 52 − 4(1)(− 5)
=3
x2 − x
x2 + x = 3 x2 − x
2(1)
(
− 5 ± 45
2
−5 ± 3 5
=
2
−5 − 3 5
−5+ 3 5
x=
or x =
2
2
≈ −5.854
≈ 0.854
Since log3 ( −8.854 + 1) = log3 ( −7.854 ) is
=
2
x + x = 3x − 3x
−2 x + 4 x = 0
−2 x ( x − 2 ) = 0
− 2 x = 0 or x − 2 = 0
x = 0 or
x=2
Since each of the original logarithms are not
defined for x = 0 , but are defined for x = 2 , the
solution set is {2} .
⎪⎧ − 5 + 3 5 ⎪⎫
⎨
⎬ ≈ {0.854} .
2
⎪⎭
⎩⎪
30. log 4 ( x 2 − 9) − log 4 ( x + 3) = 3
⎛ x2 − 9 ⎞
log 4 ⎜⎜
⎟⎟ = 3
⎝ x+3 ⎠
( x − 3)( x + 3)
= 43
x+3
x − 3 = 64
x = 67
Since each of the original logarithms is defined
for x = 67 , the solution set is {67} .
28. log 2 ( x + 1) + log 2 ( x + 7 ) = 3
log 2 ⎡⎣( x + 1)( x + 7 ) ⎤⎦ = 3
( x + 1)( x + 7 ) = 23
x2 + 7 x + x + 7 = 8
x2 + 8x − 1 = 0
− 8 ± 82 − 4(1)(−1)
2(1)
−8 ± 68
2
− 8 ± 2 17
=
2
= − 4 ± 17
31. log a ( x − 1) − log a ( x + 6) = log a ( x − 2) − log a ( x + 3)
=
⎛ x −1 ⎞
⎛ x−2⎞
log a ⎜
⎟ = log a ⎜
⎟
⎝ x+6⎠
⎝ x+3⎠
x −1 x − 2
=
x+6 x+3
( x − 1)( x + 3) = ( x − 2 )( x + 6 )
x = − 4 − 17 or x = − 4 + 17
≈ −8.123
≈ 0.123
Since log 2 ( −8.123 + 1) = log 2 ( −7.123) is
x 2 + 2 x − 3 = x 2 + 4 x − 12
2 x − 3 = 4 x − 12
9 = 2x
9
x=
2
Since each of the original logarithms is defined for
9
⎧9 ⎫
x = , the solution set is ⎨ ⎬ .
2
⎩2⎭
undefined, the solution set is
{− 4 +
)
2
2
undefined, the solution set is
x=
−1
}
17 ≈ {0.123} .
661
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
32. log a x + log a ( x − 2) = log a ( x + 4)
38. 2− x = 1.5
− x = log 2 1.5
log a ( x ( x − 2) ) = log a ( x + 4)
x( x − 2) = x + 4
x = − log 2 1.5 = −
2
x − 2x = x + 4
The solution set is
⎧ log1.5 ⎫
{− log 2 1.5} = ⎨−
⎬ ≈ {− 0.585} .
⎩ log 2 ⎭
2
x − 3x − 4 = 0
( x − 4)( x + 1) = 0
x = 4 or x = −1
Since log a (−1) is undefined, the solution set is
( )
39. 5 23 x = 8
{4} .
33. 2
x −5
23 x =
=8
2 = 23
x −5 = 3
x =8
The solution set is {8} .
34. 5− x = 25
5− x = 5 2
−x = 2
x = −2
The solution set is {−2} .
(
40.2 x =
ln10
≈ 3.322
ln 2
The solution set is
ln10 ⎫
{log 2 10} = ⎧⎨
⎬ ≈ {3.322} .
⎩ ln 2 ⎭
36. 3x = 14
ln14
≈ 2.402
ln 3
The solution set is
ln14 ⎫
{log3 14} = ⎧⎨
⎬ ≈ {2.402} .
⎩ ln 3 ⎭
41.
x = log3 14 =
log 8
2
3
⎛2⎞
0.2 x = log 4 ⎜ ⎟
⎝3⎠
log 4 (2 / 3) ln ( 2 / 3)
=
≈ −1.462
x=
0.2
0.2 ln 4
The solution set
⎧ log (2 / 3) ⎫ ⎧⎪ ln ( 2 / 3) ⎫⎪
is ⎨ 4
⎬=⎨
⎬ ≈ {−1.462} .
⎩ 0.2
⎭ ⎪⎩ 0.2 ln 4 ⎪⎭
x = log 2 10 =
log (1.2 )
)
40. 0.3 40.2 x = 0.2
35. 2 x = 10
x = − log8 1.2 = −
8
5
⎛8⎞
3 x = log 2 ⎜ ⎟
⎝5⎠
1
⎛ 8 ⎞ ln ( 8 / 5 )
≈ 0.226
x = log 2 ⎜ ⎟ =
3
3ln 2
⎝5⎠
The solution set is
⎧1
⎛ 8 ⎞ ⎫ ⎪⎧ ln ( 8 / 5 ) ⎪⎫
⎨ log 2 ⎜ ⎟ ⎬ = ⎨
⎬ ≈ {0.226} .
⎝ 5 ⎠ ⎭ ⎩⎪ 3ln 2 ⎭⎪
⎩3
x −5
37. 8− x = 1.2
− x = log8 1.2
log1.5
≈ − 0.585
log 2
(
31− 2 x = 4 x
)
( )
ln 31− 2 x = ln 4 x
(1 − 2 x) ln 3 = x ln 4
ln 3 − 2 x ln 3 = x ln 4
ln 3 = 2 x ln 3 + x ln 4
ln 3 = x(2 ln 3 + ln 4)
ln 3
≈ 0.307
x=
2 ln 3 + ln 4
ln 3
⎧
⎫
The solution set is ⎨
⎬ ≈ {0.307} .
⎩ 2 ln 3 + ln 4 ⎭
≈ − 0.088
The solution set is
⎧ log (1.2 ) ⎪⎫
{− log8 1.2} = ⎪⎨
⎬ ≈ {− 0.088} .
⎪⎩ − log 8 ⎪⎭
662
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
42.
(
2 x +1 = 51− 2 x
)
(
ln 2 x +1 = ln 51− 2 x
)
ln1.2 x = ln(0.5) − x
x ln (1.2 ) = − x ln ( 0.5 )
( x + 1) ln 2 = (1 − 2 x) ln 5
x ln 2 + ln 2 = ln 5 − 2 x ln 5
x ln 2 + 2 x ln 5 = ln 5 − ln 2
x(ln 2 + 2 ln 5) = ln 5 − ln 2
ln 5 − ln 2
≈ 0.234
x=
ln 2 + 2 ln 5
⎧ ln 5 − ln 2 ⎫
The solution set is ⎨
⎬ ≈ {0.234} .
⎩ ln 2 + 2 ln 5 ⎭
x ln (1.2 ) + x ln ( 0.5 ) = 0
x ( ln (1.2 ) + ln ( 0.5 ) ) = 0
x=0
The solution set is {0} .
46.
(
x ( ln ( 0.3) − 2 ln (1.7 ) ) = − ln (1.7 ) − ln ( 0.3)
x=
x ln ( 3 / 5 ) + x ln 7 = ln 7
ln 7
≈ 1.356
ln ( 3 / 5 ) + ln 7
47.
⎧⎪
⎫⎪
ln 7
The solution set is ⎨
⎬ ≈ {1.356} .
⎪⎩ ln ( 3 / 5 ) + ln 7 ⎪⎭
= 5x
( )
⎛4⎞
ln ⎜ ⎟ = ln 5 x
⎝3⎠
(1 − x) ln ( 4 / 3) = x ln 5
ln ( 4 / 3) − x ln ( 4 / 3) = x ln 5
ln ( 4 / 3) = x ln 5 + x ln ( 4 / 3)
48.
ln ( 4 / 3) = x ( ln 5 + ln ( 4 / 3) )
x=
ln ( 4 / 3)
ln 5 + ln ( 4 / 3)
ln ( 0.3) − 2 ln (1.7 )
≈ − 0.297
π1− x = e x
ln π1− x = ln e x
(1 − x) ln π = x
ln π − x ln π = x
ln π = x + x ln π
ln π = x(1 + ln π)
ln π
≈ 0.534
x=
1 + ln π
⎧ ln π ⎫
The solution set is ⎨
⎬ ≈ {0.534} .
⎩1 + ln π ⎭
1− x
1− x
− ln (1.7 ) − ln ( 0.3)
The solution set is
⎪⎧ − ln (1.7 ) − ln ( 0.3) ⎪⎫
⎨
⎬ ≈ {− 0.297} .
⎩⎪ ln ( 0.3) − 2 ln (1.7 ) ⎭⎪
x ( ln ( 3 / 5 ) + ln 7 ) = ln 7
⎛4⎞
⎜ ⎟
⎝3⎠
)
x ln ( 0.3) − 2 x ln (1.7 ) = − ln (1.7 ) − ln ( 0.3)
)
x ln ( 3 / 5 ) = ln 7 − x ln 7
44.
(
ln ( 0.3) + x ln ( 0.3) = 2 x ln (1.7 ) − ln (1.7 )
⎛3⎞
ln ⎜ ⎟ = ln 71− x
⎝5⎠
x ln ( 3 / 5 ) = (1 − x) ln 7
x=
)
(1 + x) ln ( 0.3) = (2 x − 1) ln (1.7 )
⎛3⎞
1− x
⎜ ⎟ =7
5
⎝ ⎠
x
(
0.31+ x = 1.7 2 x −1
ln 0.31+ x = ln 1.7 2 x −1
x
43.
1.2 x = (0.5) − x
45.
e x +3 = π x
ln e x + 3 = ln π x
x + 3 = x ln π
3 = x ln π − x
3 = x (ln π − 1)
3
x=
≈ 20.728
ln π − 1
⎧ 3 ⎫
The solution set is ⎨
⎬ ≈ {20.728} .
⎩ ln π − 1 ⎭
≈ 0.152
⎧⎪ ln ( 4 / 3) ⎫⎪
The solution set is ⎨
⎬ ≈ {0.152} .
⎪⎩ ln 5 + ln ( 4 / 3) ⎪⎭
663
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
53. 16 x + 4 x +1 − 3 = 0
22 x + 2 x − 12 = 0
49.
(2 )
x 2
(2
x
(4 )
(4 )
+ 2 x − 12 = 0
)(
)
− 3 2x + 4 = 0
2x − 3 = 0
or
2x = 3
or
( )
ln 2 x = ln 3
2x + 4 = 0
2x = − 4
No solution
x 2
x
x
2 (1)
(
x = log 4 −2 + 7
(we ignore the first solution since 4 is never
negative)
)
{ (
The solution set is log 4 −2 + 7
3x = − 2
No solution
(3 )
(3 )
3x = 1
x=0
or
3x = − 4
No solution
x
− 2 2x + 6 = 0
)(
x
)
2 −2 = 0
or
2x = 2
x =1
or
− 3 ⋅ 3x + 1 = 0
3±
( −3)2 − 4 (1)(1) 3 ±
=
2 (1)
2
5
Therefore, we get
3± 5
3x =
2
⎛ 3± 5 ⎞
x = log3 ⎜⎜
⎟⎟
⎝ 2 ⎠
The solution set is
⎧⎪
⎛ 3− 5 ⎞
⎛ 3 + 5 ⎞ ⎫⎪
⎟⎟ , log 3 ⎜⎜
⎟⎟ ⎬
⎨log3 ⎜⎜
⎪⎩
⎝ 2 ⎠
⎝ 2 ⎠ ⎭⎪
22 x + 2 x + 2 − 12 = 0
+ 22 ⋅ 2 x − 12 = 0
x 2
u=
The solution set is {0} .
x 2
− 3 ⋅ 3x + 1 = 0
u 2 − 3u + 1 = 0
a = 1, b = −3, c = 1
x
3x + 4 = 0
2 x
Let u = 3x .
x
or
)} ≈ {−0.315} .
9 x − 3x +1 + 1 = 0
54.
(3 ) + 3 ⋅ 3 − 4 = 0
(3 − 1)(3 + 4 ) = 0
3x − 1 = 0
)
x
32 x + 3x +1 − 4 = 0
x
−4 ± 28
2
4 x = −2 − 7 or 4 x = −2 + 7
The solution set is {0} .
x 2
=
−4 ± 2 7
= −2 ± 7
2
Therefore, we get
+2 =0
3x = 1 or
x=0
(2 )
(2
−4 ± 42 − 4 (1)( −3)
=
3x − 1 = 0 or 3x + 2 = 0
52.
+ 4 ⋅ 4x − 3 = 0
+ 3x − 2 = 0
(3 − 1)(3
51.
x 2
u=
32 x + 3x − 2 = 0
(3 )
+ 4 ⋅ 4x − 3 = 0
Let u = 4 x .
u 2 + 4u − 3 = 0
a = 1, b = 4, c = −3
x ln 2 = ln 3
ln 3
x=
≈ 1.585
ln 2
⎧ ln 3
⎫
The solution set is ⎨
≈ 1.585⎬ .
ln
2
⎩
⎭
50.
2 x
2x + 6 = 0
2x = − 6
No solution
≈ {−0.876, 0.876} .
The solution set is {1} .
664
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
55.
58. 2 ⋅ 49 x + 11 ⋅ 7 x + 5 = 0
25 x − 8 ⋅ 5 x = −16
(5 )
(5 )
2 x
− 8 ⋅ 5 x = −16
x 2
− 8 ⋅ 5 x = −16
( )
2 ⋅ (7 )
u 2 − 8u + 16 = 0
u=4
Therefore, we get
5x = 4
x = log 5 4
The solution set is {log5 4} ≈ {0.861} .
36 x − 6 ⋅ 6 x = −9
2 x
x 2
− 6 ⋅ 6x + 9 = 0
− 6⋅6 + 9 = 0
x
−3
)
2
(4 )
x 2
=0
x 2
(4 ) − 3⋅ 4
( 4 − 5)( 4
The solution set is {log 6 3} ≈ {0.613} .
x
57. 3 ⋅ 4 + 4 ⋅ 2 + 8 = 0
( )
3⋅(2 )
3⋅ 2
2 x
+ 4 ⋅ 2x + 8 = 0
x 2
+ 4 ⋅ 2x + 8 = 0
− 10 = 3 ⋅ 4 x
x 2
x
− 10 = 0
x
x
+2 =0
)
or 4 x + 2 = 0
x
4 −5 = 0
4x = 5
x = log 4 5
4 x = −2
The solution set is {log 4 5} ≈ {1.161} .
Let u = 2 x .
3u 2 + 4u + 8 = 0
a = 3, b = 4, c = 8
u=
− 10 ⋅ 4− x ⋅ 4 x = 3 ⋅ 4 x
(4 )
6x = 3
x = log 6 3
x
+ 11 ⋅ 7 x + 5 = 0
59. 4 x − 10 ⋅ 4− x = 3
Multiply both sides of the equation by 4 x .
x
(6
x 2
2u + 1 = 0
or u + 5 = 0
2u = −1
u = −5
1
u=−
2
Therefore, we get
1
7x = −
or 7 x = −5
2
Since 7 x > 0 for all x, the equation has no real
solution.
( u − 4 )2 = 0
(6 )
(6 )
+ 11 ⋅ 7 x + 5 = 0
Let u = 7 x .
2u 2 + 11u + 5 = 0
( 2u + 1)( u + 5) = 0
Let u = 5 x .
u 2 − 8u = −16
56.
x
2 ⋅ 72
60. 3x − 14 ⋅ 3− x = 5
Multiply both sides of the equation by 3x .
(3 )
x 2
−4 ± 42 − 4 ( 3)( 8 )
2 ( 3)
− 14 ⋅ 3− x ⋅ 3x = 5 ⋅ 3x
(3 )
x 2
−4 ± −80
= not real
6
The equation has no real solution.
=
(3 ) − 5 ⋅ 3
( 3 − 7 )(3
− 14 = 5 ⋅ 3x
x 2
x
− 14 = 0
x
x
+2 =0
x
3 −7 = 0
3x = 7
x = log3 7
)
or 3x + 2 = 0
3x = −2
The solution set is {log 3 7} ≈ {1.771} .
665
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
61. log5 ( x + 1) − log 4 ( x − 2) = 1
Using INTERSECT to solve:
y1 = ln( x + 1) / ln(5) − ln( x − 2) / ln(4)
y2 = 1
65. e x = x 2
Using INTERSECT to solve:
y1 = e x ; y2 = x 2
3
4
0
–3
5
3
–1
–4
Thus, x ≈ −0.70 , so the solution set is {−0.70} .
Thus, x ≈ 2.79 , so the solution set is {2.79} .
66. e x = x 3
Using INTERSECT to solve:
y1 = e x ; y2 = x3
62. log 2 ( x − 1) − log 6 ( x + 2) = 2
Using INTERSECT to solve:
y1 = ln( x − 1) / ln(2) − ln( x + 2) / ln(6)
y2 = 2
12
120
4
–3
15
0
3
-3
–4
Thus, x ≈ 1.86 or x ≈ 4.54 , so the solution set is
{1.86, 4.54} .
–4
Thus, x ≈ 12.15 , so the solution set is {12.15} .
67. ln x = − x
Using INTERSECT to solve:
y1 = ln x; y2 = − x
63. e x = − x
Using INTERSECT to solve: y1 = e x ; y2 = − x
2
2
–2
3
–3
–2
Thus, x ≈ 0.57 , so the solution set is {0.57} .
68. ln(2 x) = − x + 2
Using INTERSECT to solve:
y1 = ln(2 x); y2 = − x + 2
64. e 2 x = x + 2
Using INTERSECT to solve:
y1 = e 2 x ; y2 = x + 2
3
4
3
0
2
–1
4
–2
Thus, x ≈ −0.57 , so the solution set is {−0.57} .
–3
-4
–3
4
2
–4
–1
Thus, x ≈ −1.98 or x ≈ 0.45 , so the solution set
is {−1.98, 0.45} .
Thus, x ≈ 1.16 , so the solution set is {1.16} .
666
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6
Section 7.4: Logarithmic and Exponential Equations
73. e − x = ln x
Using INTERSECT to solve:
y1 = e − x ; y2 = ln x
69. ln x = x3 − 1
Using INTERSECT to solve:
y1 = ln x; y2 = x3 − 1
2
2
2
–2
4
–2
–2
–2
4
4
–2
–2
Thus, x ≈ 1.31 , so the solution set is {1.31} .
Thus, x ≈ 0.39 or x = 1 , so the solution set is
{0.39, 1} .
74. e − x = − ln x
Using INTERSECT to solve:
y1 = e − x ; y2 = − ln x
70. ln x = − x 2
Using INTERSECT to solve:
y1 = ln x; y2 = − x 2
2
2
–2
–1
2
2
–2
Thus, x ≈ 0.57 , so the solution set is {0.57}
–6
Thus, x ≈ 0.65 , so the solution set is {0.65} .
75.
log 2 ( x + 1) − log 4 x = 1
log 2 x
=1
log 2 4
log 2 x
log 2 ( x + 1) −
=1
2
2 log 2 ( x + 1) − log 2 x = 2
log 2 ( x + 1) −
71. e x + ln x = 4
Using INTERSECT to solve:
y1 = e x + ln x; y2 = 4
5
log 2 ( x + 1) 2 − log 2 x = 2
–2
⎛ ( x + 1) 2 ⎞
log 2 ⎜⎜
⎟⎟ = 2
⎝ x ⎠
( x + 1) 2
= 22
x
x2 + 2 x + 1 = 4 x
4
–2
Thus, x ≈ 1.32 , so the solution set is {1.32} .
72. e x − ln x = 4
Using INTERSECT to solve:
y1 = e x − ln x; y2 = 4
2
–1
( x − 1) 2 = 0
x −1 = 0
x =1
Since each of the original logarithms is defined
for x = 1 , the solution set is {1} .
6
6
–1
x2 − 2x + 1 = 0
–1
2
–1
Thus, x ≈ 0.05 or x ≈ 1.48 , so the solution set is
{0.05, 1.48} .
667
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
76.
log 2 (3 x + 2) − log 4 x = 3
log 2 x
=3
log 2 (3 x + 2) −
log 2 4
log 2 x
=3
log 2 (3 x + 2) −
2
2 log 2 (3 x + 2) − log 2 x = 6
78.
log 2 (3 x + 2)2 − log 2 x = 6
⎛ (3 x + 2) 2 ⎞
log 2 ⎜⎜
⎟⎟ = 6
x
⎝
⎠
(3x + 2) 2
= 26
x
9 x 2 + 12 x + 4 = 64 x
x = 34 = 81
Since each of the original logarithms is defined
for x = 81 , the solution set is {81} .
79.
2
9 x − 52 x + 4 = 0
x=
52 ±
2− x
2
( 2 ) = 2x
2− x
2
(2 ) = 2x
3
1/ 3
( −52 )2 − 4 ( 9 )( 4 )
2 (9)
1 (2− x)
2
23
= 2x
1
(2 − x) = x 2
3
2 − x = 3x 2
52 ± 2560
18
52 ± 16 10
=
18
26 ± 8 10
=
9
≈ 5.700 or 0.078
Since each of the original logarithms is defined
for x = 0.078 and x = 5.700 , the solution set is
⎧⎪ 26 − 8 10 26 + 8 10 ⎫⎪
,
⎨
⎬ ≈ {0.078, 5.700} .
9
9
⎩⎪
⎭⎪
=
77.
log 9 x + 3log 3 x = 14
log 3 x
+ 3log 3 x = 14
log3 9
log 3 x
+ 3log 3 x = 14
2
7
log 3 x = 14
2
log 3 x = 4
3x 2 + x − 2 = 0
(3x − 2)( x + 1) = 0
2
x=
or x = −1
3
⎧ 2⎫
The solution set is ⎨−1, ⎬ .
⎩ 3⎭
80.
log16 x + log 4 x + log 2 x = 7
log 2 x log 2 x
+
+ log 2 x = 7
log 2 16 log 2 4
log 2 x log 2 x
+
+ log 2 x = 7
4
2
log 2 x + 2 log 2 x + 4 log 2 x = 28
7 log 2 x = 28
log 2 x = 4
log x
log 2 x 2 = 4
log 2 x ⋅ log 2 x = 4
( log 2 x )2 = 4
log 2 x = − 2 or log 2 x = 2
x = 2−2 or x = 22
1
or x = 4
x=
4
Since each of the original logarithms is defined
1
for both x = and x = 4 , the solution set is
4
⎧1 ⎫
⎨ , 4⎬ .
⎩4 ⎭
x = 24 = 16
Since each of the original logarithms is defined
for x = 16 , the solution set is {16} .
668
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
e x + e− x
=1
2
e x + e− x = 2
81.
84.
(
e x ( e x + e − x ) = 2e x
e
x
( e ) − 2e + 1 = 0
x
ex = 1
x=0
The solution set is {0} .
− 4 ± 20
2
−4± 2 5
=
= −2± 5
2
x = ln −2 − 5 or x = ln −2 + 5
(
{ (
is ln −2 + 5
( e x ) 2 − 6e x + 1 = 0
6 ± (−6) 2 − 4(1)(1)
2(1)
85.
(
or x = ln 3 + 2 2
)
x ≈ −1.763
or x ≈ 1.763
The solution set is
1
1
1
+
ln 5 ln 3
(ln 5)(ln 3)
ln x =
ln 3 + ln 5
(ln 5)(ln 3)
ln x =
ln15
e x − e− x
=2
2
e x − e− x = 4
⎛ (ln 5)(ln 3) ⎞
⎜
⎟
e x (e x − e− x ) = 4e x
x = e⎝ ln15 ⎠ ≈ 1.921
⎧ ⎛⎜ (ln 5)(ln 3) ⎞⎟ ⎫
The solution set is ⎨⎩e⎝ ln15 ⎠ ⎬⎭ ≈ {1.921} .
e 2 x − 1 = 4e x
(e x ) 2 − 4e x − 1 = 0
ex =
log5 x + log3 x = 1
ln x =
{ln (3 − 2 2 ) , ln (3 + 2 2 )} ≈ {−1.763,1.763} .
83.
)} ≈ {−1.444}.
ln x ln x
+
=1
ln 5 ln 3
1
1
( ln x ) ⎛⎜ + ⎞⎟ = 1
⎝ ln 5 ln 3 ⎠
6 ± 32 6 ± 4 2
=
= 3± 2 2
2
2
)
(
Since ln ( −4.236 ) is undefined, the solution set
e 2 x + 1 = 6e x
(
)
x ≈ ln ( −4.236 ) or x ≈ −1.444
e x ( e x + e − x ) = 6e x
x = ln 3 − 2 2
− 4 ± 42 − 4(1)(−1)
2(1)
=
e x + e− x
=3
2
e x + e− x = 6
=
− 1 = − 4e x
ex =
ex −1 = 0
ex =
2x
( e x ) 2 + 4e x − 1 = 0
2
(e − 1) = 0
82.
)
e x e x − e− x = − 4e x
e 2 x + 1 = 2e x
x 2
e x − e− x
= −2
2
e x − e− x = − 4
−(− 4) ± (− 4) 2 − 4(1)(−1)
2(1)
4 ± 20 4 ± 2 5
=
= 2± 5
2
2
x = ln 2 − 5 or x = ln 2 + 5
=
(
)
(
)
x ≈ ln(−0.236) or x ≈ 1.444
Since ln(−0.236) is undefined, the solution set is
{ln ( 2 + 5 )} ≈ {1.444}.
669
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
)
Chapter 7: Exponential and Logarithmic Functions
log 2 x + log 6 x = 3
86.
d.
log 2 ⎡⎣( x + 3)( 3x + 1) ⎤⎦ = 7
( x + 3)( 3x + 1) = 27
3 x 2 + 10 x + 3 = 128
3
ln x =
3x 2 + 10 x − 125 = 0
( 3x + 25 )( x − 5) = 0
1
1
+
ln 2 ln 6
3(ln 2)(ln 6)
ln x =
ln 6 + ln 2
3(ln 2)(ln 6)
ln x =
ln12
3x + 25 = 0
or x − 5 = 0
x=5
3 x = −25
25
x=−
3
⎛ 3(ln 2)(ln 6) ⎞
⎜
⎟
e⎝ ln12 ⎠
The solution set is {5} .
≈ 4.479
x=
⎧ ⎛⎜ 3(ln 2)(ln 6) ⎞⎟ ⎫
The solution set is ⎨⎩e⎝ ln12 ⎠ ⎬⎭ ≈ {4.479} .
87. a.
f ( x) + g ( x) = 7
log 2 ( x + 3) + log 2 ( 3 x + 1) = 7
ln x ln x
+
=3
ln 2 ln 6
1
1
( ln x ) ⎛⎜ + ⎞⎟ = 3
⎝ ln 2 ln 6 ⎠
e.
f ( x) − g ( x) = 2
log 2 ( x + 3) − log 2 ( 3x + 1) = 2
f ( x) = 3
log 2 ( x + 3) = 3
log 2
x + 3 = 23
x+3=8
x=5
The solution set is {5}. The point ( 5,3) is
x + 3 = 12 x + 4
−1 = 11x
1
− =x
11
⎧ 1⎫
The solution set is ⎨− ⎬ .
⎩ 11 ⎭
on the graph of f.
b.
g ( x) = 4
log 2 ( 3x + 1) = 4
3 x + 1 = 24
3 x + 1 = 16
3 x = 15
x=5
The solution set is {5}. The point ( 5, 4 ) is
88. a.
f ( x) = 2
log3 ( x + 5 ) = 2
x + 5 = 32
x+5 = 9
x=4
The solution set is {4}. The point ( 4, 2 ) is
on the graph of g.
c.
x+3
=2
3x + 1
x+3
= 22
3x + 1
x + 3 = 4 ( 3x + 1)
f ( x) = g ( x)
log 2 ( x + 3) = log 2 ( 3x + 1)
on the graph of f.
x + 3 = 3x + 1
2 = 2x
1= x
The solution set is {1}, so the graphs
intersect when x = 1 . That is, at the point
(1, 2 ) .
b.
g ( x) = 3
log3 ( x − 1) = 3
x − 1 = 33
x − 1 = 27
x = 28
The solution set is {28}. The point ( 28,3)
is on the graph of g.
670
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
f ( x) = g ( x)
c.
b.
log3 ( x + 5 ) = log 3 ( x − 1)
x + 5 = x −1
5 = −1 False
This is a contradiction, so the equation has
no solution. The graphs do not intersect.
3x +1 = 2 x + 2
( )
)
( x + 1) ln 3 = ( x + 2 ) ln 2
x ln 3 + ln 3 = x ln 2 + 2 ln 2
x ln 3 − x ln 2 = 2 ln 2 − ln 3
x ( ln 3 − ln 2 ) = 2 ln 2 − ln 3
log3 ( x + 5 ) + log3 ( x − 1) = 3
2 ln 2 − ln 3
≈ 0.710
ln 3 − ln 2
⎛ 2 ln 2 − ln 3 ⎞
f⎜
⎟ ≈ 6.541
⎝ ln 3 − ln 2 ⎠
The intersection point is roughly
( 0.710, 6.541) .
x=
log 3 ⎣⎡( x + 5 )( x − 1) ⎦⎤ = 3
( x + 5)( x − 1) = 33
x 2 + 4 x − 5 = 27
x 2 + 4 x − 32 = 0
( x + 8 )( x − 4 ) = 0
c.
x + 8 = 0 or x − 4 = 0
x = −8
x=4
Based on the graph, f ( x ) > g ( x ) for
x > 0.710 . The solution set is
{ x | x > 0.710} or ( 0.710, ∞ ) .
The solution set is {4} .
90. a.
f ( x) − g ( x) = 2
e.
(
ln 3x +1 = ln 2 x + 2
f ( x) + g ( x) = 3
d.
f ( x) = g ( x)
log3 ( x + 5 ) − log3 ( x − 1) = 2
log3
x+5
=2
x −1
x+5
= 32
x −1
x + 5 = 9 ( x − 1)
x + 5 = 9x − 9
14 = 8 x
7
=x
4
⎧7 ⎫
The solution set is ⎨ ⎬ .
⎩4⎭
b.
5 x −1 = 2 x +1
( )
x ln 5 − ln 5 = x ln 2 + ln 2
x ln 5 − x ln 2 = ln 5 + ln 2
x ( ln 5 − ln 2 ) = ln 5 + ln 2
f ( x ) = 3 x +1
x=
8
(0.710, 6.541)
ln 5 + ln 2
≈ 2.513
ln 5 − ln 2
⎛ ln 5 + ln 2 ⎞
f⎜
⎟ ≈ 11.416
⎝ ln 5 − ln 2 ⎠
The intersection point is roughly
( 2.513,11.416 ) .
4
2
)
( x − 1) ln 5 = ( x + 1) ln 2
g ( x ) = 2 x+ 2
−2
(
ln 5 x −1 = ln 2 x +1
89. a.
y
f ( x) = g ( x)
x
671
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
c.
Based on the graph, f ( x ) > g ( x ) for
93. a., b.
x > 2.513 . The solution set is
{ x | x > 2.513} or ( 2.513, ∞ ) .
y
f (x ) = 2 x +1
4
91. a., b.
⎛1
⎞
⎜ ,2 2 ⎟
⎠
⎝2
2
g (x ) = 2− x +2
−2
x
f ( x) = g ( x)
c.
2 x +1 = 2− x + 2
x +1 = −x + 2
2x = 1
1
x=
2
1
⎛ ⎞
f ⎜ ⎟ = 21/ 2 +1 = 23/ 2 = 2 2
⎝2⎠
⎛1
⎞
The intersection point is ⎜ , 2 2 ⎟ .
2
⎝
⎠
f ( x) = g ( x)
c.
2
3x = 10
x = log3 10
The intersection point is ( log 3 10,10 ) .
92. a., b.
94. a., b.
c.
f ( x) = g ( x)
2 x = 12
x = log 2 12
c.
f ( x) = g ( x)
3− x +1 = 3x − 2
−x +1 = x − 2
−2 x = −3
3
x=
2
The intersection point is ( log 2 12,12 ) .
1
3
⎛3⎞
=
f ⎜ ⎟ = 3−3/ 2 +1 = 3−1/ 2 =
3
3
⎝2⎠
⎛3 3⎞
The intersection point is ⎜⎜ ,
⎟⎟ .
⎝2 3 ⎠
672
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.4: Logarithmic and Exponential Equations
95. a.
f ( x ) = 2x − 4
b.
x
298 (1.009 )
t − 2006
= 360
360
298
t − 2006
⎛ 360 ⎞
ln (1.009 )
= ln ⎜
⎟
⎝ 298 ⎠
⎛ 180 ⎞
(t − 2006) ln (1.009 ) = ln ⎜
⎟
⎝ 149 ⎠
ln (180 /149 )
t − 2006 =
ln (1.009 )
Using the graph of y = 2 , shift the graph
down 4 units.
(1.009 )t − 2006 =
t=
+ 2006
ln (1.009 )
≈ 2027.10
According to the model, the population of the
U.S. will reach 360 million people in the
beginning of the year 2027.
b. Based on the graph, f ( x ) < 0 when x < 2 .
The solution set is { x | x < 2} or ( −∞, 2 ) .
96. a.
ln (180 /149 )
g ( x ) = 3x − 9
98. a.
x
Using the graph of y = 3 , shift the graph
down nine units.
6.53 (1.0114 )
t − 2006
= 9.25
9.25
6.53
t − 2006
⎛ 9.25 ⎞
ln (1.0114 )
= ln ⎜
⎟
⎝ 6.53 ⎠
(t − 2006) ln (1.0114 ) = ln ( 9.25 / 6.53)
(1.0114 )t − 2006 =
t − 2006 =
t=
t − 2006
b.
310
=
(1.009 )
298
t − 2006
⎛ 310 ⎞
ln (1.009 )
= ln ⎜
⎟
⎝ 298 ⎠
⎛ 155 ⎞
(t − 2006) ln (1.009 ) = ln ⎜
⎟
⎝ 149 ⎠
ln (155 /149 )
t − 2006 =
ln (1.009 )
t − 2006
t=
ln (1.0114 )
+ 2006
world will reach 9.25 billion people late in the
year 2036.
= 310
ln (155 /149 )
ln ( 9.25 / 6.53)
According to the model, the population of the
The solution set is { x | x > 2} or ( 2, ∞ ) .
298 (1.009 )
ln (1.0114 )
≈ 2036.72
b. Based on the graph, g ( x ) > 0 when x > 2 .
97. a.
ln ( 9.25 / 6.53)
t − 2006
6.53 (1.0114 )
= 11.75
11.75
6.53
t − 2006
⎛ 11.75 ⎞
ln (1.0114 )
= ln ⎜
⎟
⎝ 6.53 ⎠
(t − 2006) ln (1.0114 ) = ln (11.75 / 6.53)
(1.0114 )t − 2006 =
t − 2006 =
t=
+ 2006
ln (1.009 )
≈ 2010.41
According to the model, the population of the
U.S. will reach 310 million people around the
middle of the year 2010.
ln (11.75 / 6.53)
ln (1.0114 )
ln (11.75 / 6.53)
ln (1.0114 )
+ 2006
≈ 2057.82
According to the model, the population of the
world will reach 11.75 billion people late in the
year 2057.
673
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
99. a.
14,512 ( 0.82 ) = 9, 000
t
100. a.
9, 000
14,512
⎛ 9, 000 ⎞
t
log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
( 0.82 )t
15, 000
19, 282
⎛ 15, 000 ⎞
t
log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
b.
14,512 ( 0.82 ) = 4, 000
t
19, 282 ( 0.84 ) = 8, 000
t
8, 000
19, 282
⎛ 8, 000 ⎞
t
log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
( 0.84 )t
4, 000
( 0.82 ) =
14,512
⎛ 4, 000 ⎞
t
log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
=
⎛ 8, 000 ⎞
t log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
log ( 8, 000 /19, 282 )
t=
≈ 5.0
log ( 0.84 )
⎛ 4, 000 ⎞
t log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
log ( 4, 000 /14,512 )
t=
log ( 0.82 )
According to the model, the car will be worth
≈ 6.5
$8,000 after about 5 years.
According to the model, the car will be worth
c.
$4,000 after about 6.5 years.
19, 282 ( 0.84 ) = 2, 000
t
2, 000
19, 282
⎛ 2, 000 ⎞
t
log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
( 0.84 )t
14,512 ( 0.82 ) = 2, 000
t
2, 000
14,512
⎛ 2, 000 ⎞
t
log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
( 0.82 )t
=
⎛ 15, 000 ⎞
t log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
log (15, 000 /19, 282 )
t=
≈ 1.4
log ( 0.84 )
According to the model, the car will be worth
$15,000 after about 1.4 years.
t
c.
t
( 0.84 )t
=
⎛ 9, 000 ⎞
t log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
log ( 9, 000 /14,512 )
t=
log ( 0.82 )
≈ 2.4
According to the model, the car will be worth
$9,000 after about 2.4 years.
b.
19, 282 ( 0.84 ) = 15, 000
=
=
⎛ 2, 000 ⎞
t log ( 0.84 ) = log ⎜
⎟
⎝ 19, 282 ⎠
log ( 2, 000 /19, 282 )
t=
≈ 13.0
log ( 0.84 )
According to the model, the car will be worth
$2,000 after about 13 years.
⎛ 2, 000 ⎞
t log ( 0.82 ) = log ⎜
⎟
⎝ 14,512 ⎠
log ( 2, 000 /14,512 )
t=
log ( 0.82 )
≈ 10.0
101. Solution A: change to exponential expression;
square root method; meaning of ± ; solve.
According to the model, the car will be worth
$2,000 after about 10 years.
Solution B: log a M r = r log a M ; divide by 2;
change to exponential expression; solve.
The power rule log a M r = r log a M only applies
when M > 0 . In this equation, M = x − 1 .
Now, x = −2 causes M = −2 − 1 = −3 . Thus, if
we use the power rule, we lose the valid solution
x = −2 .
674
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.5: Compound Interest
12. P = $100, r = 0.12, t = 3.75
Section 7.5
A = Pe r t = 100e(0.12)(3.75) ≈ $156.83
1. P = $500, r = 0.06, t = 6 months = 0.5 year
I = Prt = (500)(0.06)(0.5) = $15.00
13. A = $100, r = 0.06, n = 12, t = 2
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
2. P = $5000, t = 9 months = 0.75 year, I = $500
500 = 5000r (0.75)
500
r=
(5000)(0.75)
2
2
40
1
=
= ⋅100% =
% = 13 %
15 15
3
3
1
The per annum interest rate was 13 % .
3
nt
⎛ 0.04 ⎞
= 100 ⎜ 1 +
⎟
4 ⎠
⎝
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
⎛ 0.06 ⎞
= 50 ⎜ 1 +
⎟
12 ⎠
⎝
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
(4)(2)
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
≈ $59.83
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
(4)(2.5)
≈ $609.50
nt
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
⎛ 0.12 ⎞
= 300 ⎜1 +
⎟
12 ⎠
⎝
(12)(1.5)
≈ $358.84
nt
≈ $59.14
− nt
( −365)(2.5)
≈ $860.72
⎛ 0.05 ⎞
= 600 ⎜ 1 +
⎟
365 ⎠
⎝
⎛ 0.06 ⎞
= 700 ⎜ 1 +
⎟
365 ⎠
⎝
− nt
( −12)(3.5)
≈ $626.61
− nt
⎛ 0.04 ⎞
= 600 ⎜1 +
⎟
4 ⎠
⎝
( −4)(2)
≈ $554.09
18. A = $300, r = 0.03, n = 365, t = 4
⎛ r⎞
P = A ⎜1 + ⎟
⎝ n⎠
(365)(3)
≈ $697.09
− nt
⎛ 0.03 ⎞
= 300 ⎜ 1 +
⎟
365 ⎠
⎝
8. P = $700, r = 0.06, n = 365, t = 2
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
( −4)(3)
17. A = $600, r = 0.04, n = 4, t = 2
7. P = $600, r = 0.05, n = 365, t = 3
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
⎛ 0.08 ⎞
= 75 ⎜ 1 +
⎟
4 ⎠
⎝
⎛ 0.07 ⎞
= 800 ⎜ 1 +
⎟
12 ⎠
⎝
6. P = $300, r = 0.12, n = 12, t = 1.5
nt
≈ $88.72
16. A = $800, r = 0.07, n = 12, t = 3.5
(12)(3)
⎛ 0.08 ⎞
= 500 ⎜ 1 +
⎟
4 ⎠
⎝
− nt
⎛ 0.06 ⎞
= 1000 ⎜ 1 +
⎟
365 ⎠
⎝
≈ $108.29
5. P = $500, r = 0.08, n = 4, t = 2.5
nt
( −12)(2)
15. A = $1000, r = 0.06, n = 365, t = 2.5
4. P = $50, r = 0.06, n = 12, t = 3
nt
⎛ 0.06 ⎞
= 100 ⎜ 1 +
⎟
12 ⎠
⎝
14. A = $75, r = 0.08, n = 4, t = 3
3. P = $100, r = 0.04, n = 4, t = 2
⎛ r⎞
A = P ⎜1 + ⎟
⎝ n⎠
− nt
( −365)(4)
≈ $266.08
19. A = $80, r = 0.09, t = 3.25
(365)(2)
≈ $789.24
P = Ae − r t = 80e( −0.09)(3.25) ≈ $59.71
20. A = $800, r = 0.08, t = 2.5
9. P = $10, r = 0.11, t = 2
P = Ae − r t = 800e( −0.08)(2.5) ≈ $654.98
A = Pe r t = 10e(0.11)(2) ≈ $12.46
21. A = $400, r = 0.10, t = 1
10. P = $40, r = 0.07, t = 3
P = Ae − r t = 400e( −0.10)(1) ≈ $361.93
A = Pe r t = 40e(0.07)(3) ≈ $49.35
22. A = $1000, r = 0.12, t = 1
11. P = $100, r = 0.10, t = 2.25
P = Ae − r t = 1000e( −0.12)(1) ≈ $886.92
A = Pe r t = 100e(0.10)(2.25) ≈ $125.23
675
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
23. 6% compounded quarterly:
⎛ 0.06 ⎞
A = 10, 000 ⎜ 1 +
⎟
4 ⎠
⎝
6 14 %
28. Suppose P dollars are invested for 1 year at 6%.
(4)(1)
Compounded monthly yields:
= $10, 613.64
(12 )(1)
⎛ 0.06 ⎞
A = P ⎜1 +
≈ 1.06168 P .
⎟
12 ⎠
⎝
The interest earned is
I = 1.06168 P − P = 0.06168 P
I = Prt
Thus,
0.06168 P = P ⋅ r ⋅1
0.06168 = r
The effective interest rate is 6.168%.
compounded annually:
A = 10, 000 (1 + 0.0625 ) = $10, 625
1
6 14 % compounded annually yields the larger
amount.
24. 9% compounded quarterly:
⎛ 0.09 ⎞
A = 10, 000 ⎜ 1 +
⎟
4 ⎠
⎝
(4)(1)
≈ $10,930.83
29. Suppose P dollars are invested for 1 year at 5%.
Compounded continuously yields:
A = Pe(0.05)(1) ≈ 1.05127 P
The interest earned is
I = 1.05127 P − P = 0.05127 P
I = Prt
Thus,
0.05127 P = P ⋅ r ⋅1
.05127 = r
The effective interest rate is 5.127%.
9 14 % compounded annually:
A = 10, 000 (1 + 0.0925 ) = $10,925
1
9% compounded quarterly yields the larger
amount.
25. 9% compounded monthly:
(12)(1)
⎛ 0.09 ⎞
A = 10, 000 ⎜ 1 +
⎟
12 ⎠
⎝
8.8% compounded daily:
= $10,938.07
30. Suppose P dollars are invested for 1 year at 6%.
Compounded continuously yields:
A = Pe(0.06)(1) ≈ 1.06184 P
The interest earned is
I = 1.06184 P − P = 0.06184 P
Thus,
I = Prt
0.06184 P = P ⋅ r ⋅1
0.06184 = r
The effective interest rate is 6.184%.
365
⎛ 0.088 ⎞
A = 10, 000 ⎜ 1 +
⎟ = $10,919.77
365 ⎠
⎝
9% compounded monthly yields the larger
amount.
26. 8% compounded semiannually:
(2)(1)
⎛ 0.08 ⎞
A = 10, 000 ⎜ 1 +
= $10,816
⎟
2 ⎠
⎝
7.9% compounded daily:
(
31. 2 P = P 1 + 1r
365
⎛ 0.079 ⎞
A = 10, 000 ⎜ 1 +
⎟ = $10,821.95
365 ⎠
⎝
7.9% compounded daily yields the larger
amount.
)
3(1)
2 P = P (1 + r )
3
2 = (1 + r )3
3
2 = 1+ r
r = 3 2 − 1 ≈ 0.25992
The required rate is 25.992%.
27. Suppose P dollars are invested for 1 year at 5%.
Compounded quarterly yields:
( 4 )(1)
⎛ 0.05 ⎞
A = P ⎜1 +
≈ 1.05095 P .
⎟
4 ⎠
⎝
The interest earned is
I = 1.05095 P − P = 0.05095 P
I = Prt
Thus,
P
0.05095 = P ⋅ r ⋅1
0.05095 = r
The effective interest rate is 5.095%.
(
32. 2 P = P 1 + 1r
)
2 P = P (1 + r )
6(1)
6
2 = (1 + r )6
6
2 = 1+ r
r = 6 2 − 1 ≈ 0.12246
The required rate is 12.246%.
676
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.5: Compound Interest
(
33. 3P = P 1 + 1r
)
3P = P (1 + r )
5(1)
b.
3 = e0.06t
ln 3 = 0.06t
ln 3
t=
≈ 18.31
0.06
It will take about 18.31 years to triple.
5
3 = (1 + r )5
5
3 = 1+ r
r = 5 3 − 1 ≈ 0.24573
The required rate is 24.573%.
34.
(
3P = P 1 + 1r
37. Since the effective interest rate is 7%, we have:
I = Prt
I = P ⋅ 0.07 ⋅1
I = 0.07 P
Thus, the amount in the account is
A = P + 0.07 P = 1.07 P
)
10(1)
3P = P (1 + r )
10
3 = (1 + r )10
10
3 = 1+ r
Let x be the required interest rate. Then,
r = 10 3 − 1 ≈ 0.11612
The required rate is 11.612%.
⎛ r⎞
1.07 P = P ⎜1 + ⎟
⎝ 4⎠
4
⎛ 0.08 ⎞
2 P = P ⎜1 +
⎟
12 ⎠
⎝
12t
⎛ 0.08 ⎞
2 = ⎜1 +
⎟
12 ⎠
⎝
12t
⎛ 0.08 ⎞
ln 2 = ln ⎜ 1 +
⎟
12 ⎠
⎝
⎛ 0.08 ⎞
ln 2 = 12t ln ⎜1 +
⎟
12 ⎠
⎝
ln 2
t=
≈ 8.69
⎛ 0.08 ⎞
12 ln ⎜ 1 +
⎟
12 ⎠
⎝
It will take about 8.69 years to double.
b.
(
38. Since the effective interest rate is 6%, we have:
I = Prt
I = P ⋅ 0.06 ⋅1
I = 0.06 P
Thus, the amount in the account is
A = P + 0.06 P = 1.06 P
2 P = Pe0.08t
Let x be the required interest rate. Then,
1.06 P = Pe( r )(1)
1.06 = er
r = ln(1.06) ≈ 0.05827
Thus, an interest rate of 5.827% compounded
continuously has an effective interest rate of 6%.
12t
⎛ 0.06 ⎞
3P = P ⎜ 1 +
⎟
12 ⎠
⎝
12 t
3 = (1.005 )
12t
39.
ln 3 = ln (1.005 )
12t
ln 3 = 12t ln (1.005 )
t=
)
Thus, an interest rate of 6.823% compounded
quarterly has an effective interest rate of 7%.
2 = e0.08t
ln 2 = 0.08t
ln 2
t=
≈ 8.66
0.08
It will take about 8.66 years to double.
36. a.
( 4 )(1)
⎛ r⎞
1.07 = ⎜ 1 + ⎟
⎝ 4⎠
r
4
1.07 = 1 +
4
r
4
1.07 − 1 =
4
r = 4 4 1.07 − 1 ≈ 0.06823
12t
35. a.
3P = Pe0.06t
ln 3
≈ 18.36
12 ln (1.005 )
It will take about 18.36 years to triple.
⎛ 0.08 ⎞
150 = 100 ⎜ 1 +
⎟
12 ⎠
⎝
12 t
1.5 ≈ (1.006667 )
ln1.5 ≈ 12t ln (1.006667 )
ln1.5
t≈
≈ 5.09
12 ln (1.006667 )
Compounded monthly, it will take about 5.09
years (or 61.02 months).
677
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
44. A = 200 (1 + 0.0125 ) ≈ $215.48
6
150 = 100e0.08 t
1.5 = e0.08 t
ln1.5 = 0.08 t
ln1.5
t=
≈ 5.07
0.08
Compounded continuously, it will take about
5.07 years (or 60.82 months).
Her bill will be $215.48 after 6 months.
45. P = 15, 000e( −0.05)(3) ≈ $12,910.62
Jerome should ask for $12,910.62.
( −12)(0.5)
⎛ 0.03 ⎞
≈ $2955.39
46. P = 3, 000 ⎜1 +
⎟
12 ⎠
⎝
John should save $2955.39.
12 t
40.
⎛ 0.10 ⎞
175 = 100 ⎜ 1 +
⎟
12 ⎠
⎝
47. A = 15(1 + 0.15)5 = 15(1.15)5 ≈ $30.17 per share
for a total of about $3017.
1.75 ≈ (1.008333)
12 t
ln1.75 ≈ 12t ln (1.008333)
t≈
48. 850, 000 = 650, 000 (1 + r )
85
3
= (1 + r )
65
85
3
= 1+ r
65
r ≈ 3 1.3077 − 1 ≈ 0.0935
The annual return is approximately 9.35%.
3
ln1.75
≈ 5.62
12 ln (1.008333)
Compounded monthly, it will take about 5.62
years (or 67.43 months).
175 = 100e0.10 t
1.75 = e0.10 t
ln1.75 = 0.10 t
ln1.75
t=
≈ 5.60
0.10
Compounded continuously, it will take about
5.60 years (or 67.15 months).
49. 5.6% compounded continuously:
A = 1000e(0.056)(1) = $1057.60
Jim will not have enough money to buy the
computer.
5.9% compounded monthly:
41. 25, 000 = 10, 000e0.06 t
12
⎛ 0.059 ⎞
A = 1000 ⎜ 1 +
⎟ = $1060.62
12 ⎠
⎝
The second bank offers the better deal.
2.5 = e0.06 t
ln 2.5 = 0.06 t
ln 2.5
t=
≈ 15.27
0.06
It will take about 15.27 years (or 15 years, 3
months).
50. 6.8% compounded continuously for 3 months:
Amount on April 1:
A = 1000e(0.068)(0.25) = $1017.15
5.25% compounded monthly for 1 month:
Amount on May 1
42. 80, 000 = 25, 000e0.07 t
3.2 = e0.07 t
ln 3.2 = 0.07 t
ln 3.2
t=
≈ 16.62
0.07
It will take about 16.62 years (or 16 years, 7
months).
⎛ 0.0525 ⎞
A = 1017.15 ⎜1 +
⎟
12 ⎠
⎝
(12)(1/12)
= $1021.60
51. Will: 9% compounded semiannually:
⎛ 0.09 ⎞
A = 2000 ⎜ 1 +
⎟
2 ⎠
⎝
43. A = 90, 000(1 + 0.03)5 = $104,335
The house will cost $104,335 in three years.
(2)(20)
= $11, 632.73
Henry: 8.5% compounded continuously:
A = 2000e(0.085)(20) = $10,947.89
Will has more money after 20 years.
678
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.5: Compound Interest
52. Value of $1000 compounded continuously at
10% for 3 years:
A = 1000e(0.10)(3) = $1349.86
56. From 2006 to 2007, the deficit decreased at a
423 − 354
rate of r =
≈ 0.16312 or 16.312%.
423
The deficit t years after 2007 is given by
April will have more money if she takes the
$1000 now and invests it.
53. a.
D = 354 (1 − 0.16312 ) = 354 ( 0.83688 ) .
t
Now, if D = 100 million, then
Let x = the year, then the average cost C of a
4-year private college is by the function
C ( x) = 29, 026(1.055) x − 2005 .
354 ( 0.83688 ) = 100
100
( 0.83688)t =
354
t
⎛ 100 ⎞
ln ( 0.83688 ) = ln ⎜
⎟
⎝ 354 ⎠
⎛ 100 ⎞
t ln ( 0.83688 ) = ln ⎜
⎟
⎝ 354 ⎠
⎛ 100 ⎞
ln ⎜
⎟
⎝ 354 ⎠ ≈ 7.1
t=
ln ( 0.83688 )
t
C (2015) = 29, 026(1.055) 2015− 2005
= 29, 026(1.055)10
≈ 49,581
In 2015, the average cost of college at a 4year private college will be about $49,581.
b.
A = Pe rt
49,581 = Pe0.04(10)
49,581
P = 0.04(10) ≈ 33, 235
e
An investment of $33,235 in 2005 would
pay for the cost of college at a 4-year private
college in 2015.
The deficit will be $100 billion approximately
7.1 years after 2007, or in the fiscal year 2014.
57. P = 1000, r = 0.03, n = 2
A = 1000(1 − 0.03) 2 = $940.90
54. P = 100, 000; t = 5
a. Simple interest at 12% per annum:
A = 100, 000 + 100, 000(0.12)(5) = $160, 000
I = $160, 000 − $100, 000 = $60, 000
58. P = 1000, r = 0.02, n = 3
A = 1000(1 − 0.02)3 ≈ $941.19
59. P = 1000, A = 950, n = 2
b. 11.5% compounded monthly:
950 = 1000(1 − r ) 2
(12)(5)
⎛ 0.115 ⎞
A = 100, 000 ⎜ 1 +
≈ $177, 227
⎟
12 ⎠
⎝
I = $177, 227 − $100, 000 = $77, 227
0.95 = (1 − r ) 2
± 0.95 = 1 − r
r = 1 ± 0.95
r ≈ 0.0253 or r ≈ 1.9747
Disregard r ≈ 1.9747 . The inflation rate was
2.53%.
11.25% compounded continuously:
A = 100, 000e(0.1125)(5) ≈ $175,505
I = $175, 505 − $100, 000 = $75,505
Thus, simple interest at 12% is the best option
since it results in the least interest.
c.
⎛ r⎞
55. A = P ⎜ 1 + ⎟
⎝ n⎠
t
60. P = 1000, A = 930, n = 2
930 = 1000(1 − r ) 2
nt
0.93 = (1 − r ) 2
± 0.93 = 1 − r
2(20)
40
⎛ 0.032 ⎞
A = 319 ⎜ 1 +
= 319 (1.016 ) ≈ 602
⎟
2 ⎠
⎝
The government would have to pay back
approximately $602 billion in 2025.
r = 1 ± 0.93
r ≈ 0.0356 or r ≈ 1.9644
Disregard r ≈ 1.9644 . The inflation rate was
3.56%.
679
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
61. r = 0.02
1
P = P(1 − 0.02)t
2
0.5 P = P(0.98)t
0.5 = (0.98)t
t = log 0.98 (0.5)
ln 0.5
=
≈ 34.31
ln 0.98
The purchasing power will be half in 34.31 years.
62. r = 0.04
1
P = P(1 − 0.04)t
2
0.5 P = P(0.96)t
t=
b.
t=
c.
b.
( −12)(20)
≈ $1364.62
68. a.
A = $10, 000, r = 0.10, t = 20
P = 10, 000e( −0.10)(20) ≈ $1353.35
≈ $4631.93
66. A = $25, 000, P = 12, 485.52, n = 1, t = 8
(
25, 000 = 12, 485.52 1 + r 8
)
25, 000
8
= (1 + r )
12, 485.52
8
ln 8000 − ln1000
≈ 20.79 years
0.10
ln 30, 000 − ln 2000
r
ln 30, 000 − ln 2000
r=
35
≈ 0.0774
r ≈ 7.74%
c.
A = Per t
A
= er t
P
⎛ A⎞
ln ⎜ ⎟ = r t
⎝P⎠
ln A − ln P = r t
ln A − ln P
t=
r
≈ $10,810.76
( −1)(10)
t=
35 =
−17
65. A = $10, 000, r = 0.08, n = 1, t = 10
⎛ 0.08 ⎞
P = 10, 000 ⎜ 1 +
⎟
1 ⎠
⎝
nt
b.
64. A = $40, 000, r = 0.08, n = 1, t = 17
⎛ 0.08 ⎞
P = 40, 000 ⎜ 1 +
⎟
1 ⎠
⎝
⎛ r⎞
mP = P ⎜ 1 + ⎟
⎝ n⎠
nt
A = $10, 000, r = 0.10, n = 12, t = 20
⎛ 0.10 ⎞
P = 10, 000 ⎜ 1 +
⎟
12 ⎠
⎝
ln 3
⎛ 0.06 ⎞
4 ⋅ ln ⎜ 1 +
⎟
4 ⎠
⎝
ln 3
=
≈ 18.45 years
4 ln (1.015 )
⎛ r⎞
m = ⎜1 + ⎟
⎝ n⎠
⎛ r⎞
ln m = n t ⋅ ln ⎜1 + ⎟
⎝ n⎠
ln m
t=
⎛ r⎞
n ⋅ ln ⎜1 + ⎟
⎝ n⎠
0.5 = (0.96)t
t = log 0.96 (0.5)
ln 0.5
=
≈ 16.98
ln 0.96
The purchasing power will be half in 16.98 years.
63. a.
ln 2
⎛ 0.12 ⎞
1 ⋅ ln ⎜ 1 +
⎟
1 ⎠
⎝
ln 2
=
≈ 6.12 years
ln (1.12 )
67. a.
25, 000
= 1+ r
12, 485.52
25, 000
−1
12, 485.52
r ≈ 0.090665741
The annual rate of return is about 9.07%.
r=8
680
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Exponential Growth and Decay; Newton’s Law; Logistic Growth and Decay Models
69. a.
CPI 0 = 152.4, CPI = 195.3,
n = 2005 − 1995 = 10
72. CPI 0 = 100, CPI = 456.5, r = 5.57
⎛ 5.57 ⎞
456.5 = 100 ⎜1 +
⎟
⎝ 100 ⎠
10
r ⎞
⎛
195.3 = 152.4 ⎜ 1 +
⎟
⎝ 100 ⎠
456.5 = 100 (1.0557 )
10
195.3 ⎛
r ⎞
= ⎜1 +
⎟
152.4 ⎝ 100 ⎠
r
195.3
= 10
1+
100
152.4
r
195.3
= 10
−1
100
152.4
⎛ 195.3 ⎞
− 1⎟⎟ ≈ 2.51%
r = 100 ⎜⎜ 10
⎝ 152.4 ⎠
b.
4.565 = (1.0557 )
300 ⎛ 2.51 ⎞
= ⎜1 +
⎟
152.4 ⎝ 100 ⎠
ln 4.565
ln1.0558
≈ 28.0 years
The yeas that was used as the base period for the
CPI was about 28 years before 1995, or the year
1967.
73 – 75. Answers will vary.
n
n
Section 7.6
1. P (t ) = 500e0.02 t
n
a.
P (0) = 500e(
0.02 )⋅ ( 0 )
= 500 insects
b. growth rate: k = 0.02 = 2 %
c.
P (10) = 500e(
0.02 )⋅ (10 )
≈ 611 insects
d. Find t when P = 800 :
800 = 500e0.02 t
1.6 = e0.02 t
ln1.6 = 0.02 t
ln1.6
t=
≈ 23.5 days
0.02
70. CPI 0 = 234.2, r = 2.8%, n = 5
5
2.8 ⎞
⎛
CPI = 234.2 ⎜ 1 +
⎟ ≈ 268.9
⎝ 100 ⎠
In 5 years, the CPI index will be about 268.9.
e.
Find t when P = 1000 :
1000 = 500e0.02 t
2 = e0.02 t
ln 2 = 0.02 t
ln 2
t=
≈ 34.7 days
0.02
71. r = 3.1%
2 = (1.031)
n
=
⎛ 300 ⎞
⎛ 2.51 ⎞
ln ⎜
⎟ = ln ⎜ 1 +
⎟
⎝ 152.4 ⎠
⎝ 100 ⎠
⎛ 300 ⎞
⎛ 2.51 ⎞
ln ⎜
⎟ = n ln ⎜1 +
⎟
⎝ 152.4 ⎠
⎝ 100 ⎠
⎛ 300 ⎞
ln ⎜
⎟
⎝ 152.4 ⎠ ≈ 27.3 years
n=
⎛ 2.51 ⎞
ln ⎜ 1 +
⎟
⎝ 100 ⎠
The CPI will reach 300 about 27 years after
1995, or in the year 2022.
3.1 ⎞
⎛
2 ⋅ CPI 0 = CPI 0 ⎜ 1 +
⎟
⎝ 100 ⎠
n
n = log1.0558 ( 4.565 )
CPI 0 = 152.4, CPI = 300, r = 2.51
⎛ 2.51 ⎞
300 = 152.4 ⎜ 1 +
⎟
⎝ 100 ⎠
n
n
n
2. N (t ) = 1000e0.01t
ln 2
≈ 22.7
ln1.031
It will take about 22.7 years for the CPI index to
double.
n = log1.031 2 =
a.
N (0) = 1000e(
0.01)⋅ ( 0 )
= 1000 bacteria
b. growth rate: k = 0.01 = 1 %
c.
N (4) = 1000e(
0.01)⋅ ( 4 )
≈ 1041 bacteria
681
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
d. Find t when N = 1700 :
d. Find t when A = 50 :
50 = 100e−0.087 t
1700 = 1000e0.01t
0.5 = e−0.087 t
ln 0.5 = − 0.087 t
ln 0.5
t=
≈ 7.97 days
− 0.087
1.7 = e0.01t
ln1.7 = 0.01t
ln1.7
t=
≈ 53.1 hours
0.01
e.
Find t when N = 2000 :
2000 = 1000e0.01t
5. a.
b. If N (t ) = 1800, N 0 = 1000, and t = 1 , then
2 = e0.01t
ln 2 = 0.01t
ln 2
t=
≈ 69.3 hours
0.01
−0.0244 t
1800 = 1000ek (1)
1.8 = ek
k = ln1.8
If t = 3 , then N (3) = 1000e(
mosquitoes.
−0.0244 t
3. A(t ) = A0 e
= 500e
a. decay rate: k = −0.0244 = −2.44%
( −0.0244 )(10 )
b.
A(10) = 500e
c.
Find t when A = 400 :
c.
t=
6. a.
N (t ) = N 0 e k t
800 = 500e k (1)
1.6 = e k
k = ln1.6
If t = 5 , then N (5) = 500e(
bacteria
c.
ln1.6 )( 5 )
≈ 5243
Find t when N (t ) = 20, 000 :
20, 000 = 500e(
−0.087 t
4. A(t ) = A0 e
= 100e
a. decay rate: k = −0.087 = −8.7%
c.
ln10
≈ 3.9 days
ln1.8
b. If N (t ) = 800, N 0 = 500, and t = 1 , then
0.5 = e−0.0244 t
ln 0.5 = − 0.0244 t
ln 0.5
t=
≈ 28.4 years
− 0.0244
−0.087 )( 9 )
ln1.8) t
ln1.8 t
d. Find t when A = 250 :
250 = 500e−0.0244 t
A(9) = 100e(
= 5832
10 = e( )
ln10 = ( ln1.8 ) t
0.8 = e−0.0244 t
ln 0.8 = −0.0244 t
ln 0.8
t=
≈ 9.1 years
−0.0244
b.
ln1.8 )( 3)
Find t when N (t ) = 10, 000 :
10, 000 = 1000e(
≈ 391.7 grams
400 = 500e−0.0244 t
−0.087 t
N (t ) = N 0 e k t
ln1.6 ) t
40 = e( )
ln 40 = ( ln1.6 ) t
ln1.6 t
≈ 45.7 grams
t=
Find t when A = 70 :
70 = 100e−0.087 t
ln 40
≈ 7.85 hours
ln1.6
0.7 = e−0.087 t
ln 0.7 = −0.087 t
ln 0.7
t=
≈ 4.1 days
−0.087
682
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Exponential Growth and Decay; Newton’s Law; Logistic Growth and Decay Models
7. a.
When A0 = 10 and t = 100 :
N (t ) = N 0 ek t
⎛ ln 0.5 ⎞
(100 )
⎜
9⎟
⎠
b. Note that 18 months = 1.5 years, so t = 1.5.
2 N 0 = N 0 ek (1.5)
A = 10e⎝ 1.3×10
When A0 = 10 and t = 1000 :
2 = e1.5k
ln 2 = 1.5k
ln 2
k=
1.5
If N 0 = 10, 000 and t = 2 , then
⎛ ln 0.5 ⎞
(1000 )
⎜
9⎟
⎠
A = 10e⎝ 1.3×10
≈ 9.999994668 grams
11. Use A = A0 ekt and solve for k :
half-life = 5600 years
⎛ ln 2 ⎞
⎜
⎟ ( 2)
P (2) = 10, 000e⎝ 1.5 ⎠ ≈ 25,198
The population 2 years from now will be
25,198.
8. a.
≈ 9.999999467 grams
0.5 A0 = A0 ek (5600)
0.5 = e5600 k
ln 0.5 = 5600k
ln 0.5
k=
5600
N (t ) = N 0 e k t , k < 0
b. If N (t ) = 800, 000, N 0 = 900, 000, and
t = 2005 − 2003 = 2 , then
800, 000 = 900, 000ek (2)
8
= e2k
9
⎛8⎞
ln ⎜ ⎟ = 2k
⎝9⎠
ln ( 8 / 9 )
k=
2
If t = 2007 − 2003 = 4 , then
Solve for t when A = 0.3 A0 :
0.3 A0 = A0
⎛ ln 0.5 ⎞
⎜
⎟t
e⎝ 5600 ⎠
⎛ ln 0.5 ⎞
⎜
⎟t
0.3 = e⎝ 5600 ⎠
⎛ ln 0.5 ⎞
ln 0.3 = ⎜
⎟t
⎝ 5600 ⎠
5600
t=
( ln 0.3) ≈ 9727
ln 0.5
The tree died approximately 9727 years ago.
⎛ ln ( 8 / 9 ) ⎞
⎜
⎟ ( 4)
2 ⎠
900, 000e⎝
P (4) =
≈ 711,111
The population in 2007 will be 711,111.
12. Use A = A0 e kt and solve for k :
half-life = 5600 years
0.5 A0 = A0 e k (5600)
9. Use A = A0 ek t and solve for k :
0.5 A0 = A0 ek (1690)
0.5 A0 = A0 e k (5600)
0.5 = e1690 k
ln 0.5 = 1690k
ln 0.5
k=
1690
When A0 = 10 and t = 50 :
Solve for t when A = 0.7 A0 :
⎛ ln 0.5 ⎞
⎜
⎟ ( 50 )
A = 10e⎝ 1690 ⎠
0.5 = e5600 k
ln 0.5 = 5600k
ln 0.5
k=
5600
⎛ ln 0.5 ⎞
⎜
⎟t
≈ 9.797 grams
0.7 A0 = A0 e⎝ 5600 ⎠
10. Use A = A0 e k t and solve for k :
0.5 A0
(
=Ae
0.5 = e
k 1.3×109
0
0.7 =
)
⎛ ln 0.5 ⎞
⎜
⎟t
e⎝ 5600 ⎠
⎛ ln 0.5 ⎞
ln 0.7 = ⎜
⎟t
⎝ 5600 ⎠
5600
t=
( ln 0.7 ) ≈ 2882
ln 0.5
The fossil is about 2882 years old.
(1.3×109 )k
ln 0.5 = 1.3 × 109 k
ln 0.5
k=
1.3 × 109
683
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
13. a.
Using u = T + (u0 − T )e k t with t = 5 ,
T = 70 , u0 = 450 , and u = 300 :
14. a.
Using u = T + (u0 − T )e kt with t = 2 ,
T = 38, u0 = 72 , and u = 60 :
300 = 70 + (450 − 70)ek (5)
60 = 38 + (72 − 38)ek (2)
22 = 34e 2 k
22
= e2k
34
⎛ 22 ⎞
ln ⎜ ⎟ = 2k
⎝ 34 ⎠
ln ( 22 / 34 )
k=
2
T = 38, u0 = 72, t = 7
230 = 380e5k
230
= e5 k
380
⎛ 23 ⎞
ln ⎜ ⎟ = 5k
⎝ 38 ⎠
1 ⎛ 23 ⎞
k = ln ⎜ ⎟ ≈ − 0.1004
5 ⎝ 38 ⎠
T = 70, u0 = 450, u = 135 :
135 =
65 =
65
=e
380
ln ( 23/ 38)
t
70 + (450 − 70)e 5
u=
⎛ ln ( 22 / 34 ) ⎞
⎜
⎟( 7 )
2
ln ( 23/ 38)
t
380e 5
⎠
u = 38 + 34e⎝
≈ 45.41º F
After 7 minutes the thermometer will read
about 45.41˚F.
b. Find t when u = 39º F
ln ( 23/ 38)
t
5
⎛ 65 ⎞ ln ( 23 / 38 )
ln ⎜
t
⎟=
5
⎝ 380 ⎠
5
⎛ 65 ⎞
t=
⋅ ln ⎜
⎟ ≈ 18 minutes
ln ( 23 / 38 ) ⎝ 380 ⎠
39 =
1=
90 =
⎛ ln ( 23/ 38) ⎞
⎜
⎟t
5
⎠
70 + (450 − 70)e⎝
t=
⎛ ln ( 23/ 38) ⎞
⎜
⎟t
5
⎠
380e⎝
⎛ ln ( 23/ 38) ⎞
⎟t
5
⎠
⎜
90
= e⎝
380
c.
2
⎛ 1 ⎞
⋅ ln ⎜ ⎟ ≈ 16.2
ln ( 22 / 34 ) ⎝ 34 ⎠
The thermometer will read 39 degrees after
about 16.2 minutes.
T = 38, u0 = 72, u = 45 :
ln ( 22 / 34 )
45 = 38 + (72 − 38)e
ln ( 23 / 38 )
⎛ 90 ⎞
ln ⎜
t
⎟=
5
⎝ 380 ⎠
5
⎛ 90 ⎞
t=
⋅ ln ⎜
⎟ ≈ 14.3 minutes
ln ( 23 / 38 ) ⎝ 380 ⎠
c.
⎛ ln ( 22 / 34 ) ⎞
⎜
⎟t
2
⎠
34e⎝
⎛ ln ( 22 / 34 ) ⎞
T = 70, u0 = 450, u = 160 :
160 =
⎛ ln ( 22 / 34 ) ⎞
⎜
⎟t
2
⎠
38 + (72 − 38)e⎝
⎜
⎟t
1
2
⎠
= e⎝
34
⎛ 1 ⎞ ⎛ ln ( 22 / 34 ) ⎞
ln ⎜ ⎟ = ⎜
⎟t
2
⎝ 34 ⎠ ⎝
⎠
The temperature of the pan will be 135˚F at
about 5:18 PM.
b.
⎛ ln ( 22 / 34 ) ⎞
⎜
⎟( 7 )
2
⎠
38 + (72 − 38)e⎝
7 = 34e
7
=e
34
The pan will be 160˚F after about 14.3
minutes.
As time passes, the temperature of the pan
approaches 70˚F.
2
t
ln ( 22 / 34 )
t
2
ln ( 22 / 34 )
2
t
⎛ 7 ⎞ ln(22 / 34)
t
ln ⎜ ⎟ =
2
⎝ 34 ⎠
2
⎛ 7 ⎞
t=
⋅ ln ⎜ ⎟ ≈ 7.26 minutes
ln(22 / 34) ⎝ 34 ⎠
The thermometer will read 45° F after
about 7.26 minutes.
d. As time passes, the temperature gets closer
to 38˚F.
684
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Exponential Growth and Decay; Newton’s Law; Logistic Growth and Decay Models
⎛ 25 ⎞ ⎛ ln ( 35 / 42 ) ⎞
ln ⎜ ⎟ = ⎜
⎟t
10
⎝ 42 ⎠ ⎝
⎠
10
⎛ 25 ⎞
t=
⋅ ln ⎜ ⎟ ≈ 28.46
ln ( 35 / 42 ) ⎝ 42 ⎠
15. Using u = T + (u0 − T )e kt with t = 3 , T = 35 ,
u0 = 8 , and u = 15 :
15 = 35 + (8 − 35)e k (3)
−20 = − 27e3k
20
= e3k
27
⎛ 20 ⎞
ln ⎜ ⎟ = 3k
⎝ 27 ⎠
ln ( 20 / 27 )
k=
3
The temperature of the stein will be 45° F after
about 28.46 minutes.
17. Use A = A0 e kt and solve for k:
2.2 = 2.5e k (24)
0.88 = e 24 k
ln 0.88 = 24k
ln 0.88
k=
24
At t = 5 :
⎛ ln ( 20 / 27 ) ⎞
⎜
⎟ ( 5)
3
⎠
35 + (8 − 35)e⎝
u=
≈ 18.63°C
After 5 minutes, the thermometer will read
approximately 18.63°C .
When A 0 = 2.5 and t = 72 :
⎛ ln 0.88 ⎞
⎜
⎟ ( 72 )
A = 2.5e⎝ 24 ⎠
≈ 1.70
After 3 days (72 hours), the amount of free
chlorine will be 1.70 parts per million.
At t = 10 :
⎛ ln ( 20 / 27 ) ⎞
⎜
⎟ (10 )
3
⎠
35 + (8 − 35)e⎝
u=
≈ 25.1°C
After 10 minutes, the thermometer will read
approximately 25.1°C
Find t when A = 1 :
⎛ ln 0.88 ⎞
⎜
⎟t
24 ⎠
1 = 2.5 e⎝
16. Using u = T + (u0 − T )e kt with t = 10 , T = 70 ,
⎛ ln 0.88 ⎞
⎜
⎟t
0.4 = e⎝ 24 ⎠
⎛ ln 0.88 ⎞
ln 0.4 = ⎜
⎟t
⎝ 24 ⎠
24
t=
⋅ ln 0.4 ≈ 172
ln 0.88
Ben will have to shock his pool again after 172
hours (or 7.17 days) when the level of free
chlorine reaches 1.0 parts per million.
u0 = 28 , and u = 35 :
35 = 70 + (28 − 70)e k (10)
− 35 = − 42e10 k
35
= e10 k
42
⎛ 35 ⎞
ln ⎜ ⎟ = 10k
⎝ 42 ⎠
ln ( 35 / 42 )
k=
10
18. Use A = A0 e kt and solve for k:
0.15 = 0.25e k (17)
At t = 30 :
u=
≈ 45.69°F
After 30 minutes, the temperature of the stein
will be approximately 45.69° F .
0.6 = e17 k
ln 0.6 = 17k
ln 0.6
k=
17
Find the value of t so that the u = 45˚F:
When A 0 = 0.25 and t = 30 :
⎛ ln ( 35 / 42 ) ⎞
⎜
⎟ ( 30 )
10
⎠
70 + (28 − 70)e⎝
45 =
− 25 =
⎛ ln ( 35/ 42 ) ⎞
⎜
⎟t
10
⎠
70 + (28 − 70)e⎝
⎛ ln 0.6 ⎞
⎜
⎟ ( 30 )
0.25e⎝ 17 ⎠
A=
≈ 0.10
After 30 minutes, approximately 0.10 M of
dinitrogen pentoxide will remain.
⎛ ln ( 35/ 42 ) ⎞
⎜
⎟t
10
⎠
− 42e⎝
⎛ ln ( 35/ 42 ) ⎞
⎟t
10
⎠
⎜
25
= e⎝
42
685
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
Find t when A = 0.01 :
0.01 =
⎛ ln 0.6 ⎞
⎜
⎟t
0.25 e⎝ 17 ⎠
0.04 =
⎛ ln 0.6 ⎞
⎜
⎟t
e⎝ 17 ⎠
20. Use A = A0 e kt and solve for k :
15 = 25e k (10)
0.6 = e10 k
ln 0.6 = 10k
ln 0.6
k=
10
⎛ ln 0.6 ⎞
ln 0.04 = ⎜
⎟t
⎝ 17 ⎠
17
t=
⋅ ln 0.04 ≈ 107
ln 0.6
It will take approximately 107 minutes until 0.01
M of dinitrogen pentoxide remains.
When A0 = 25 and t = 24 :
⎛ ln 0.6 ⎞
⎜
⎟ ( 24 )
25e⎝ 10 ⎠
A=
≈ 7.34
There will be about 7.34 kilograms of salt left
after 1 day.
19. Use A = A0 e kt and solve for k:
Find t when A = 0.5 A0 :
0.36 = 0.40e k (30)
30 k
0.9 = e
ln 0.9 = 30k
ln 0.9
k=
30
0.5 =
⎛ ln 0.6 ⎞
⎜
⎟t
0.02 = e⎝ 10 ⎠
⎛ ln 0.6 ⎞
ln 0.02 = ⎜
⎟t
⎝ 10 ⎠
10
t=
⋅ ln 0.02 ≈ 76.6
ln 0.6
It will take about 76.6 hours until ½ kilogram of
salt is left.
Note that 2 hours = 120 minutes.
When A 0 = 0.40 and t = 120 :
⎛ ln 0.9 ⎞
⎜
⎟ (120 )
A = 0.40e⎝ 30 ⎠
≈ 0.26
After 2 hours, approximately 0.26 M of sucrose
will remain.
21. Use A = A0 e kt and solve for k :
Find t when A = 0.10 :
0.10 =
⎛ ln 0.6 ⎞
⎜
⎟t
25 e⎝ 10 ⎠
0.5 A0 = A0 e k (8)
⎛ ln 0.9 ⎞
⎜
⎟t
0.40e⎝ 30 ⎠
0.5 = e8 k
ln 0.5 = 8k
ln 0.5
k=
8
⎛ ln 0.9 ⎞
⎜
⎟t
0.25 = e⎝ 30 ⎠
⎛ ln 0.9 ⎞
ln 0.25 = ⎜
⎟t
⎝ 30 ⎠
30
t=
⋅ ln 0.25 ≈ 395
ln 0.9
It will take approximately 395 minutes (or 6.58
hours) until 0.10 M of sucrose remains.
Find t when A = 0.1A0 :
⎛ ln 0.5 ⎞
⎜
⎟t
8 ⎠
0.1A0 = A0 e⎝
0.1 =
⎛ ln 0.5 ⎞
⎜
⎟t
e⎝ 8 ⎠
⎛ ln 0.5 ⎞
ln 0.1 = ⎜
⎟t
⎝ 8 ⎠
8
t=
⋅ ln 0.1 ≈ 26.6
ln 0.5
The farmers need to wait about 26.6 days before
using the hay.
686
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.6: Exponential Growth and Decay; Newton’s Law; Logistic Growth and Decay Models
24. a.
22. Using u = T + (u0 − T )e kt with t = 2 , T = 325 ,
u0 = 75 , and u = 100 :
100 = 325 + (75 − 325)ek (2)
b.
−225 = − 250e2 k
0.9 = e2 k
2k = ln 0.9
ln 0.9
k=
2
Find the value of t so that u = 175˚F:
175 =
−150 =
0.6 =
c.
⎛ ln 0.9 ⎞
⎜
⎟t
325 + (75 − 325)e⎝ 2 ⎠
⎛ ln 0.9 ⎞
⎜
⎟t
− 250e⎝ 2 ⎠
⎛ ln 0.9 ⎞
⎜
⎟t
e⎝ 2 ⎠
⎛ ln 0.9 ⎞
ln 0.6 = ⎜
⎟t
⎝ 2 ⎠
2
t=
⋅ ln 0.6 ≈ 9.7
ln 0.9
The hotel may serve their guests about 9.7 hours
after noon or at about 9:42 PM.
23. a.
b.
c.
0.9
=
0.90
1 + 3.5e−0.339(4)
0.90
=
≈ 0.473
1 + 3.5e−1.356
After 4 months, Intel’s latest coprocessor
will be in about 47.3% of PCs sold at Best
Buy.
t=
ln ( 2 / 35 )
≈ 8.44
−0.339
75% of PCs sold at Best Buy have Intel’s
latest coprocessor about 8.44 months after it
has been introduced.
t = 2005 − 2000 = 5
0.9
0.9
=
≈ 0.4070
P (5) =
−0.32(5)
1 + 6e
1 + 6e−1.6
In 2005, about 40.7% of U.S. households
owned a DVD player.
e.
d. We need to find t such that P = 0.8.
0.9
0.8 =
1 + 6e−0.32 t
9
1 + 6e−0.32t =
8
1
−0.32t
6e
=
8
1
e−0.32t =
48
−0.32t = ln (1/ 48 )
t=
P (4) =
3.5e−0.339t = 0.2
2
e−0.339t =
35
−0.339t = ln ( 2 / 35 )
0.9
0.9
=
= 0.1286
1 + 6 ⋅1 7
1 + 6e
In 2000, about 12.86% of U.S. households
owned a DVD player.
−0.32(0)
0.90
0.90
=
1 + 3.5e−0.339(0) 1 + 3.5 ⋅1
0.90
=
= 0.2
4.5
When it is first introduced, Intel’s latest
coprocessor will be in 20% of computers
sold at Best Buy.
P (0) =
d. We need to find t such that P = 0.75.
0.90
0.75 =
1 + 3.5e−0.339 t
1 + 3.5e−0.339t = 1.2
The maximum proportion is the carrying
capacity, c = 0.9 = 90%.
P (0) =
The maximum proportion is the carrying
capacity, c = 0.90 = 90%.
We need to find t such that P = 0.45.
0.90
0.45 =
1 + 3.5e−0.339 t
1 + 3.5e−0.339t = 2
3.5e−0.339t = 1
2
e−0.339t =
7
−0.339t = ln ( 2 / 7 )
t=
ln ( 2 / 7 )
≈ 3.70
−0.339
45% of PCs sold at Best Buy have Intel’s
latest coprocessor about 3.7 months after it
has been introduced.
ln (1/ 48 )
≈ 12.1
−0.32
Since 2000 + 12.1 = 2012.1 , 80% of
households will own a DVD player in 2012.
687
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
25. a.
As t → ∞, e −0.439 t → 0. Thus, P(t ) → 1000 .
The carrying capacity is 1000 grams of
bacteria.
26. a.
As t → ∞, e−0.162 t → 0. Thus, P(t ) → 500 .
The carrying capacity is 500 bald eagles.
b. Growth rate = 0.162 = 16.2%.
b. Growth rate = 0.439 = 43.9%.
c.
c.
1000
1000
P (0) =
=
= 30
−0.439(0)
33.33
1 + 32.33e
The initial population was 30 grams of
bacteria.
P (9) =
≈ 616.6
1 + 32.33e−0.439(9)
After 9 hours, the population of bacteria will
be about 616.6 grams.
e.
We need to find t such that P = 700:
1000
700 =
1 + 32.33e−0.439 t
10
1 + 32.33e−0.439t =
7
3
32.33e−0.439t =
7
3
−0.439t
e
=
226.31
−0.439t = ln ( 3 / 226.31)
t=
ln ( 3 / 226.31)
−0.439
e.
≈ 9.85
e −0.162t ≈ 0.012
−0.162t = ln ( 0.012 )
t=
ln ( 0.012 )
−0.162
≈ 27.3
The bald eagle population will reach onehalf of its carrying capacity after about 27.3
years.
1
32.33
−0.439t = ln (1/ 32.33)
e−0.439t =
−0.439
We need to find t such that
1
P = ( 500 ) = 250 :
2
500
250 =
1 + 83.33e−0.162 t
−0.162t
1 + 83.33e
=2
83.33e −0.162t = 1
32.33e−0.439t = 1
t=
ln ( 0.008 )
≈ 29.8
−0.162
The bald eagle population will be 300 in
approximately 29.8 years.
We need to find t such that
1
P = (1000 ) = 500 :
2
1000
500 =
1 + 32.33e −0.439 t
−0.439t
1 + 32.33e
=2
ln (1/ 32.33)
≈ 9.56
1 + 83.33e −0.162(3)
After 3 years, the population is almost 10
bald eagles.
t=
The population of bacteria will be 700
grams after about 9.85 hours.
f.
500
d. We need to find t such that P = 300:
500
300 =
1 + 83.33e−0.162 t
5
1 + 83.33e −0.162t =
3
2
83.33e −0.162t =
3
e −0.162t ≈ 0.008
−0.162t = ln ( 0.008 )
1000
d.
P (3) =
27. a.
≈ 7.9
The population of bacteria will reach onehalf of is carrying capacity after about 7.9
hours.
b.
y=
6
≈ 0.00923
1+ e
At 100° F , the predicted number of eroded
or leaky primary O-rings will be about 0.
y=
− (5.085 − 0.1156(100))
6
≈ 0.81
1+ e
At 60° F , the predicted number of eroded or
leaky primary O-rings will be about 1.
− (5.085 − 0.1156(60))
688
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.7: Building Exponential, Logarithmic, and Logistic Models from Data
c.
d.
Section 7.7
6
≈ 5.01
1 + e− (5.085− 0.1156(30))
At 30°F , the predicted number of eroded or
leaky primary O-rings will be about 5.
y=
Y1 =
1
1. a.
6
1+ e
− (5.085 − 0.1156 x )
−1
6
7
0
b. Using EXPonential REGression on the data
yields: y = 0.0903 (1.3384 )
c.
0
100
0
Use INTERSECT with Y2 = 1, 3, and 5 :
6
y = 0.0903 (1.3384 )
x
x
(
ln 1.3384 )
= 0.0903 e (
)
x
ln 1.3384 ) x
= 0.0903e (
N ( t ) = 0.0903e0.2915t
d.
0
Y1 = 0.0903e0.2915 x
1
100
0
The predicted number of eroded or leaky
O-rings is 1 when the temperature is about
57.91°F .
6
0
100
0
100
−1
7
0
e.
N ( 7 ) = 0.0903e(
f.
We need to find t when N = 0.75 :
0.0903e(
0.2915 )⋅7
≈ 0.69 bacteria
0.2915 )⋅t
= 0.75
0.75
0.2915 )⋅t
e(
=
0.0903
⎛ 0.75 ⎞
0.2915t = ln ⎜
⎟
⎝ 0.0903 ⎠
⎛ 0.75 ⎞
ln ⎜
⎟
0.0903 ⎠
≈ 7.26 hours
t≈ ⎝
0.2915
0
The predicted number of eroded or leaky
O-rings is 3 when the temperature is about
43.99°F .
6
1.5
2. a.
0
The predicted number of eroded or leaky
O-rings is 5 when the temperature is about
30.07°F .
−1
0
6
b. Using EXPonential REGression on the data
yields: y = 0.0339 (1.9474 )
x
689
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
c.
y = 0.0339 (1.9474 )
x
(
ln 1.9474 )
= 0.0339 e (
e.
)
We need to find t when A ( t ) = 0.5 ⋅ A0
100.3263e(
x
d.
Y1 = 0.0339e(
e
= 0.5
−0.1314t = ln 0.5
ln 0.5
t=
≈ 5.3 weeks
−0.1314
0.6665 )t
0.6665 ) x
1.5
f.
A ( 50 ) = 100.3263e(
g.
We need to find t when A ( t ) = 20 .
100.3263e(
−1
0
0.6665 )⋅6
≈ 0.14 grams
−0.1314 )t
( −0.1314 )t
e.
N ( 6 ) = 0.0339e(
f.
We need to find t when N = 2.1 :
≈ 1.85 bacteria
0.6665 )t
20 ⎞
⎛
ln ⎜
⎟
100.3263 ⎠
t= ⎝
≈ 12.3 weeks
−0.1314
= 2.1
2.1
( 0.6665)t
e
=
0.0339
⎛ 2.1 ⎞
0.6665t = ln ⎜
⎟
⎝ 0.0339 ⎠
⎛ 2.1 ⎞
ln ⎜
⎟
0.0339 ⎠
≈ 6.19 hours
t≈ ⎝
0.6665
1100
4. a.
−1
110
3. a.
−0.1314 )⋅50
= 20
20
e
=
100.3263
20 ⎞
⎛
−0.1314t = ln ⎜
⎟
100.3263
⎝
⎠
6
0.0339e(
= ( 0.5 )(100.3263)
( −0.1314 )t
ln 1.9474 ) x
= 0.0339e (
N ( t ) = 0.0339e(
−0.1314 )t
8
400
b. Using EXPonential REGression on the data
yields: y = 998.907 ( 0.8976 )
−1
7
40
c.
c.
y = 100.3263 ( 0.8769 )
(
x
(
)
A ( t ) = 998.907e(
x
d.
ln 0.8769 ) x
= 100.3263e (
A ( t ) = 100.3263e(
d.
Y1 = 100.3263e(
40
x
Y1 = 998.907e(
−0.1080 )t
−0.1080 ) x
1100
−0.1314 )t
−0.1314 ) x
110
−1
)
ln 0.8976 ) x
= 998.907e (
x
ln 0.8769 )
= 100.3263 e (
x
ln 0.8976 )
= 998.907 e (
b. Using EXPonential REGression on the data
yields: y = 100.3263 ( 0.8769 )
y = 998.907 ( 0.8976 )
x
−1
400
8
7
690
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.7: Building Exponential, Logarithmic, and Logistic Models from Data
e.
We need to find t when A ( t ) = 0.5 ⋅ A0
998.907e(
−0.1080 )t
e.
A (15 ) = 751.470e(
= ( 0.5 )( 998.907 )
−0.1080 t
A ( 20 ) = 998.907e(
g.
We need to find t when A = 200 :
998.907e(
−0.1080 )⋅20
f.
≈ 359 billion
We need to find t when A ( t ) = 230 .
751.470e(
−0.04932 )t
= 230
230
e
=
751.470
⎛ 230 ⎞
−0.04932t = ln ⎜
⎟
⎝ 751.470 ⎠
⎛ 230 ⎞
ln ⎜
⎟
751.470 ⎠
≈ 24 years
t= ⎝
−0.04932
Now 1995 + 24 = 2019. The number of
cigarettes produced in the U.S. will decrease to
230 billion in the year 2019.
( −0.04932 )t
≈ 115 grams
−0.1080 )t
= 200
200
−0.1080 )t
e(
=
998.907
⎛ 200 ⎞
−0.1080t = ln ⎜
⎟
⎝ 998.907 ⎠
⎛ 200 ⎞
ln ⎜
⎟
998.907 ⎠
t= ⎝
≈ 14.9 days
−0.1080
5. a.
−0.04932 )⋅15
cigarettes.
)
e(
= 0.5
−0.1080t = ln 0.5
ln 0.5
t=
≈ 6.42 days
−0.1080
f.
Note that 2010 is represented by t = 15 .
6. a.
Let x = 0 correspond to 1995, x = 3
correspond to 1998, etc.
250
Let x = 0 correspond to 1995, x = 3
correspond to 1998, etc.
800
−1
10
100
b. Using EXPonential REGression on the data
−1
yields: y = 228.4370 ( 0.92301)
10
450
c.
b. Using EXPonential REGression on the data
yields: y = 751.4698 ( 0.95188 )
c.
y = 751.4698 ( 0.95188 )
(
= 751.4698e
Y1 = 751.470e(
A ( t ) = 228.437e(
)
x
d.
Y1 = 228.437e(
)
x
−0.08012 )t
−0.08012 )t
250
ln ( 0.95188 ) x
A ( t ) = 751.470e(
d.
ln 0.92301) x
= 228.4370e (
x
x
ln 0.95188 )
= 751.4698 e (
(
ln 0.92301)
y = 228.4370 e (
x
−0.04932 )t
−0.04932 ) x
−1
800
e.
100
10
Note that 2010 is represented by t = 15 .
A (15 ) = 228.437e(
−0.08012 )⋅15
≈ 68.7 billion
cigarettes.
−1
450
10
691
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
f.
We need to find t when A ( t ) = 230 .
228.437e(
−0.08012 )t
= 50
50
228.437
⎛ 50 ⎞
−0.08012t = ln ⎜
⎟
⎝ 228.437 ⎠
⎛ 50 ⎞
ln ⎜
⎟
228.437 ⎠
t= ⎝
≈ 19 years
−0.08012
Now 1995 + 19 = 2014. The number of
cigarettes exported from the U.S. will decrease
to 50 billion in the year 2014.
e
( −0.08012 )t
=
110
900
190
b. Using LnREGression on the data
yields: y = −11,850.72 + 2688.50 ln x
c.
Y1 = −11,850.72 + 2688.50 ln x
2400
2400
7. a.
110
150
900
190
2400
9. a.
900
190
e5.0216 ≈ x
x ≈ 152
If the price were $1650, then approximately
152 computers would be supplied.
Y1 = 32, 741.02 − 6070.96 ln x
150
900
d. Find x when y = 1650:
1650 = −11,850.72 + 2688.50 ln x
13,500.72 = 2688.50 ln x
13,500.72
= ln x
2688.50
5.0216 ≈ ln x
b. Using LnREGression on the data yields:
y = 32, 741.02 − 6070.96 ln x
c.
2400
8. a.
190
d. We need to find x when y = 1650 :
1650 = 32, 741.02 − 6070.96 ln x
−31, 091.02 = −6070.96 ln x
−31, 091.02
= ln x
−6070.96
5.1213 ≈ ln x
Let x = 0 correspond to 1900, x = 10
correspond to 1910, x = 20 correspond to
1920, etc.
300,000,000
−10
70,000,000
110
b. Using LOGISTIC REGression on the data
799, 475,916.5
yields: y =
1 + 9.1968e−0.01603 x
e5.1213 ≈ x
x ≈ 168
If the price were $1650, then approximately
168 computers would be demanded.
c.
Y1 =
799, 475,916.5
1 + 9.1968e−0.01603 x
300,000,000
−10
70,000,000
110
692
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Section 7.7: Building Exponential, Logarithmic, and Logistic Models from Data
d. As x → ∞ , 9.1968e −0.01603 x → 0 , which
means 1 + 9.1968e −0.01603 x → 1 , so
799, 475,916.5
y=
→ 799, 475,916.5
1 + 9.1968e −0.01603 x
Therefore, the carrying capacity of the
United States is approximately 799,475,917
people.
e.
The year 2004 corresponds to x = 104, so
799, 475,916.5
y=
−0.01603(104 )
1 + 9.1968e
≈ 292,177,932 people
f.
Find x when y = 300, 000, 000
799, 475,916.5
= 300, 000, 000
1 + 9.1968e−0.01603 x
(
799, 475, 916.5 = 300, 000, 000 1 + 9.1968e
−0.01603 x
c.
7
0
)
f.
We need to find x when y = 7 :
)
10.05267
= 1 + 0.8439e −0.0320 x
7
10.05267
− 1 = 0.8439e −0.0320 x
7
0.4361 ≈ 0.8439e −0.0320 x
0.4361
≈ e −0.0320 x
0.8439
⎛ 0.4361 ⎞
ln ⎜
⎟ ≈ −0.0320 x
⎝ 0.8439 ⎠
⎛ 0.4361 ⎞
ln ⎜
⎟
0.8439 ⎠
x≈ ⎝
≈ 21
−0.0320
Therefore, the world population will be 7
billion in approximately the year 2013.
7
5
10
5
(
Let x = 1 correspond to 1993, x = 2
correspond to 1994, etc.
0
:
10.05267
=7
1 + 0.8439e −0.0320 x
10.05267 = 7 1 + 0.8439e −0.0320 x
−0.01603 x
1.6649
≈ e −0.01603 x
9.1968
⎛ 1.6649 ⎞
ln ⎜
⎟ ≈ −0.01603 x
⎝ 9.1968 ⎠
⎛ 1.6649 ⎞
ln ⎜
⎟
⎝ 9.1968 ⎠ ≈ x
−0.01603
x ≈ 107
Therefore, the United States population will
be 300,000,000 around the year 2007.
10. a.
10.05267
1 + 0.8439e −0.0320 x
d. As x → ∞ , 0.8439e −0.0320 x → 0 , which
means 1 + 0.8439e −0.0320 x → 1 , so
10.05267
y=
→ 10.05267
1 + 0.8439e −0.0320 x
Therefore, the carrying capacity of the world
is approximately 10.053 billion people.
e. The year 2004 corresponds to x = 12, so
10.05267
y=
≈ 6.38 .
−0.0320(12 )
1 + 0.8439e
In 2004, the population of the world was
approximately 6.38 billion people.
799, 475,916.5
= 1 + 9.1968e −0.01603 x
300, 000, 000
799, 475,916.5
− 1 = 9.1968e −0.01603 x
300, 000, 000
1.6649 ≈ 9.1968e
Y1 =
11. a.
10
b. Using LOGISTIC REGression on the data
10.05267
yields: y =
1 + 0.8439e −0.0320 x
Let x = 5 correspond to 1975, x = 10
correspond to 1980, x = 20 correspond to
1990, etc.
80,000
0
0
40
693
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
d. The year 2009 corresponds to x = 25 , so
268.9893
y=
≈ 245.04 .
1 + 219.0605e −0.3086(25)
For 2009, the function predicts that the
number of cell phone subscribers will be
approximately 245,040,000 subscribers.
b. Using LOGISTIC REGression on the data
68, 684.7826
yields: y =
1 + 18.9416e−0.19743 x
c.
Y1 =
68, 684.7826
1 + 18.9416e−0.19743 x
80,000
0
e.
40
0
d. As x → ∞ , 18.9416e −0.19743 x → 0 , which
means 1 + 18.9416e −0.19743 x → 1 , so
68, 684.7826
y=
→ 68, 684.7826
1 + 18.9416e −0.19743 x
Therefore, the carrying capacity of the cable
TV market in the U.S. is about 68,685,000
subscribers.
e.
12. a.
b.
Chapter 7 Review Exercises
1. a.
The year 2015 corresponds to x = 45 , so
68, 684.7826
≈ 68,505 .
y=
1 + 18.9416e−0.19743(45)
In 2015, cable TV will have approximately
68,505,000 subscribers in the U.S.
Let x = 1 correspond to 1985, x = 2
correspond to 1986, x = 3 correspond to
1980, etc. Using LOGISTIC REGression on
268.9893
the data yields: y =
1 + 219.0605e −0.3086 x
Y1 =
The answer in part (e) of Example 1 predicts
1,512,010,000 cell phone subscribers in the
U.S. in 2009. This prediction is more than 6
times our prediction in part (d). The
function in Example 1 assumes exponential
growth which is unlimited growth. Our
logistic function in part (d) assumes that the
growth is limited.
g ( 9 ) = log3 ( 9 ) = log 3 32 = 2
c.
f ( −2 ) = 3−2 =
d.
⎛ 1 ⎞
⎛ 1 ⎞
g ⎜ ⎟ = log3 ⎜ ⎟ = log3 3−3 = −3
27
⎝ ⎠
⎝ 27 ⎠
1
9
( )
f (1) = 31 = 3
( )
b.
g ( 81) = log 3 ( 81) = log3 34 = 4
c.
f ( −4 ) = 3−4 =
d.
⎛ 1 ⎞
⎛ 1 ⎞
−5
g⎜
⎟ = log 3 ⎜
⎟ = log3 3 = −5
⎝ 243 ⎠
⎝ 243 ⎠
1 + 219.0605e −0.3086 x
240
( )
b.
2. a.
268.9893
f ( 4 ) = 34 = 81
1
81
( )
3. 52 = z is equivalent to 2 = log 5 z
−1
23
4. a 5 = m is equivalent to 5 = log a m
−20
c.
As x → ∞ , 219.0605e −0.3086 x → 0 , which
means 1 + 219.0605e −0.3086 x → 1 , so
268.9893
y=
→ 268.9893
1 + 219.0605e −0.3086 x
Therefore, the carrying capacity of the
cellular phone market in the U.S. is about
268,989,300 subscribers.
5. log 5 u = 13 is equivalent to 513 = u
6. log a 4 = 3 is equivalent to a 3 = 4
694
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
7.
f ( x ) = log(3x − 2) requires:
12. log3 81 = log 3 34 = 4 log3 3 = 4
3x − 2 > 0
2
x>
3
⎧
2⎫
⎛2 ⎞
Domain: ⎨ x x > ⎬ or ⎜ , ∞ ⎟
3⎭
⎝3 ⎠
⎩
15. 2log 20.4 = 0.4
16. log 2 2
2x + 1 > 0
)
9. H ( x ) = log 2 x 2 − 3 x + 2 requires
(
p( x) = x − 3 x + 2 > 0
( x − 2 )( x − 1) > 0
(2, ∞)
(1, 2)
(
0
(
Thus, the domain of H ( x ) = log 2 x 2 − 3 x + 2
(
= 4 log 2 a 2 b
(
)
)
(
)
)
19. log x 2 x3 + 1 = log x 2 + log x3 + 1
1/ 2
(
)
1
= 2 log x + log x 3 + 1
2
)
⎛ x2 + 2 x + 1 ⎞
2
2
20. log5 ⎜⎜
⎟⎟ = log 5 ( x + 1) − log5 x
2
x
⎝
⎠
= 2 log 5 ( x + 1) − 2 log 5 x
( )
is { x x < 1 or x > 2} or ( −∞,1) ∪ ( 2, ∞ ) .
)
10. F ( x ) = ln x 2 − 9 requires
⎛ x 3 x2 + 1 ⎞
⎟ = ln x 3 x 2 + 1 − ln( x − 3)
21. ln ⎜
⎜ x−3 ⎟
⎝
⎠
)
(
2
p ( x) = x − 9 > 0
( x + 3)( x − 3) > 0
(
Interval
(−∞, −3)
(−3,3)
(3, ∞ )
Test Value
−4
0
4
Value of p
7
−9
7
Conclusion
positive
(
1/ 3
− ln( x − 3)
)
1
= ln x + ln x 2 + 1 − ln( x − 3)
3
2
negative positive
(
)
= ln x + ln x 2 + 1
x = −3 and x = 3 are the zeros of p .
22.
)
Thus, the domain of F ( x ) = ln x 2 − 9 is
{x
4
1
⎛
⎞
= 4 ⎜ 2 log 2 a + log 2 b ⎟
2
⎝
⎠
= 8log 2 a + 2 log 2 b
3
3
2
1
−
Value of p
2
2
4
Conclusion positive negative positive
(
)
= 4 log 2 a 2 + log 2 b1/ 2
x = 2 and x = 1 are the zeros of p .
Test Value
= 3 log 2 2 = 3
18. log 2 a 2 b
2
(−∞,1)
3
⎛ uv 2 ⎞
2
17. log3 ⎜⎜
⎟⎟ = log 3 uv − log 3 w
w
⎝
⎠
= log 3 u + log 3 v 2 − log 3 w
= log 3 u + 2 log 3 v − log 3 w
1
x>−
2
⎧
1⎫
⎛ 1 ⎞
Domain: ⎨ x x > − ⎬ or ⎜ − , ∞ ⎟
2⎭
⎝ 2 ⎠
⎩
Interval
= 2
14. eln 0.1 = 0.1
8. F ( x ) = log 5 (2 x + 1) requires:
(
2
13. ln e
x < −3 or x > 3} or ( −∞, 3) ∪ ( 3, ∞ ) .
⎛1⎞
11. log 2 ⎜ ⎟ = log 2 2−3 = −3log 2 2 = −3
⎝8⎠
⎛ 2x + 3 ⎞
ln ⎜ 2
⎟
⎝ x − 3x + 2 ⎠
⎛ 2x + 3 ⎞
= 2 ln ⎜ 2
⎟
⎝ x − 3x + 2 ⎠
= 2 ( ln(2 x + 3) − ln [ ( x − 1)( x − 2) ])
= 2 ( ln(2 x + 3) − ln( x − 1) − ln( x − 2) )
= 2 ln(2 x + 3) − 2 ln( x − 1) − 2 ln( x − 2)
695
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
23.
( )
( )
3
1
3log 4 x 2 + log 4 x = log 4 x 2 + log 4 x1/ 2
2
= log 4 x 6 + log 4 x1/ 4
(
= log 4 x 6 ⋅ x1/ 4
1
[log( x + 3) + log( x − 2)]
2
1
= log 22 + log x3 − log [ ( x + 3)( x − 2) ]
2
1/ 2
27. 2 log 2 + 3log x −
)
( )
= log 4 x3 − log ( ( x + 3)( x − 2) )
25 / 4
= log 4 x
25
=
log 4 x
4
⎛
4 x3
= log ⎜
⎜ [ ( x + 3)( x − 2) ]1/ 2
⎝
⎛1⎞ 1
24. − 2 log3 ⎜ ⎟ + log3 x
⎝ x⎠ 3
( )
= log 3 x
−1 −2
28.
( )
1/ 2 1/ 3
+ log3 x
(
)
(
)
⎛ x −1 ⎞
⎛ x ⎞
2
25. ln ⎜
⎟ + ln ⎜
⎟ − ln x − 1
1
x
x
+
⎝
⎠
⎝
⎠
⎛ x −1 x ⎞
2
= ln ⎜
⋅
⎟ − ln x − 1
⎝ x x +1⎠
⎡ x −1 ⎤
⎢
⎥
= ln ⎢ x2 + 1 ⎥
⎣ x − 1⎦
⎛ x −1
⎞
1
= ln ⎜
⋅
⎟
+
1
(
−
1)(
+
1)
x
x
x
⎝
⎠
1
= ln
( x + 1) 2
(
(
)
)
(
)
29. log 4 19 =
ln19
≈ 2.124
ln 4
30. log 2 21 =
ln 21
≈ 4.392
ln 2
31. Y1 = log3 x =
3
ln x
ln 3
8
−1
= ln( x + 1)−2
= − 2 ln( x + 1)
26. log x 2 − 9 − log x 2 + 7 x + 12
4
1/ 2
(
= log 3 x
13
= log 3 x
6
)
)
1/ 2
⎛1⎞
− ln ⎜ ⎟ − ln ( x( x − 4) )
⎝2⎠
1/ 2
⎛
⎞
x2 + 1
⎜
⎟
= ln ⎜
⎟
⎜⎜ 1 [ x( x − 4) ]1/ 2 ⎟⎟
⎝ 16
⎠
⎛ 16 x 2 + 1 ⎞
⎟
= ln ⎜
⎜ x( x − 4) ⎟
⎝
⎠
13/ 6
(
(
⎞
⎟
⎟
⎠
1
1 1
ln x 2 + 1 − 4 ln − [ ln( x − 4) + ln x ]
2
2 2
= ln x 2 + 1
= log 3 x 2 + log 3 x1/ 6
= log 3 x 2 ⋅ x1/ 6
1/ 2
−3
)
32. Y1 = log 7 x =
⎛ ( x − 3)( x + 3) ⎞
= log ⎜
⎟
⎝ ( x + 3)( x + 4) ⎠
⎛ x−3⎞
= log ⎜
⎟
⎝ x+4⎠
3
−1
ln x
ln 7
8
−3
696
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
33.
f ( x) = 2 x −3
a.
34.
Domain: (−∞, ∞)
a.
b. Using the graph of y = 2 x , shift the graph
horizontally 3 units to the right.
c.
f ( x) = − 2 x + 3
Domain: (−∞, ∞)
b. Using the graph of y = 2 x , reflect the graph
about the x-axis, and shift vertically 3 units
up.
Range: (0, ∞)
Horizontal Asymptote: y = 0
c.
f ( x ) = 2 x −3
d.
y = 2 x −3
d.
x = 2 y −3
Inverse
y − 3 = log 2 x
y = 3 + log 2 x
f
−1
Range of f: (0, ∞)
f.
Using the graph of y = log 2 x , shift the
graph vertically 3 units up.
f ( x) = − 2 x + 3
y = − 2x + 3
x = − 2y + 3
x −3 = −2
( x) = 3 + log 2 x
e.
Range: (−∞,3)
Horizontal Asymptote: y = 3
Inverse
y
3 − x = 2y
y = log 2 ( 3 − x )
f −1 ( x ) = log 2 ( 3 − x )
e.
3− x > 0
− x > −3
x<3
Range of f: (−∞,3)
f.
Using the graph of y = log 2 x , reflect the
graph about the y-axis, and shift horizontally
to the right 3 units.
697
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
35.
f ( x) =
a.
( )
1 −x
3
2
36.
f ( x) = 1 + 3− x
a.
Domain: (−∞, ∞)
Domain: (−∞, ∞)
b. Using the graph of y = 3x , reflect the graph
about the y-axis, and shift vertically 1 unit
up.
y
b. Using the graph of y = 3x , reflect the graph
about the y-axis, and compress vertically by
1
a factor of .
2
(−1, 4)
5
(0, 2)
y=1
5 x
−5
−5
c.
d.
Range: (0, ∞)
Horizontal Asymptote: y = 0
c.
( )
( )
( )
d.
1
f ( x ) = 3− x
2
1 −x
y= 3
2
1 −y
Inverse
x= 3
2
2 x = 3− y
− y = log3 (2 x)
y = − log 3 (2 x)
f.
x = 1 + 3− y
Inverse
−y
x −1 = 3
− y = log3 ( x − 1)
y = − log 3 ( x − 1)
f
2x > 0
x>0
Range of f: (0, ∞)
Using the graph of y = log3 x , compress the
graph horizontally by a factor of
f ( x ) = 1 + 3− x
y = 1 + 3− x
f −1 ( x) = − log 3 (2 x)
e.
Range: (1, ∞)
Horizontal Asymptote: y = 1
1
, and
2
−1
( x) = − log 3 ( x − 1)
e.
x −1 > 0
x >1
Range of f: (1, ∞)
f.
Using the graph of y = log3 x , shift the
graph horizontally to the right 1 unit, and
reflect vertically about the x-axis.
y x=1
5
reflect about the x-axis.
(2, 0)
5 x
(4,−1)
−5
−5
698
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
f ( x) = 1 − e− x
37.
a.
38.
Domain: (−∞, ∞)
f ( x) = 3e x − 2
a.
b. Using the graph of y = e x , reflect about the
y-axis, reflect about the x-axis, and shift up 1
unit.
y
Domain: (−∞, ∞)
b. Using the graph of y = e x , shift the graph
two units horizontally to the right, and
stretch vertically by a factor of 3.
y
5
5
(2, 3)
y=1
(0, 0.41)
−5
(−1,−1.72)
(0, 0)
5 x
−5
−5
c.
d.
c.
Range: (−∞, 1)
Horizontal Asymptote: y = 1
d.
f ( x) = 1 − e− x
x = 1 − e− y
Range: (0, ∞)
Horizontal Asymptote: y = 0
f ( x ) = 3e x − 2
y = 3e x − 2
y = 1 − e− x
x = 3e y − 2
x
= e y −2
3
y − 2 = ln 3x
Inverse
x − 1 = −e − y
1 − x = e− y
− y = ln (1 − x )
y = − ln (1 − x )
f −1 ( x) = − ln (1 − x )
e.
f.
5 x
−5
f −1
1− x > 0
− x > −1
x <1
Range of f: (−∞,1)
Using the graph of y = ln x , reflect the
graph about the y-axis, shift to the right 1
unit, and reflect about the x-axis.
y
Inverse
( )
y = 2 + ln ( 3x )
( x ) = 2 + ln ( 3x )
e.
Range of f: (0, ∞)
f.
Using the graph of y = ln x , stretch
horizontally by a factor of 3, and shift
vertically up 2 units.
y
5
(3, 2)
5
5 x
−5
(0.41, 0)
(0, 0)
5 x
−5
−5
(−1.72,−1)
−5
x=1
699
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
39.
f ( x) =
a.
40.
1
ln ( x + 3)
2
f ( x) = 3 + ln(2 x)
a.
Domain: (−3, ∞)
Domain: (0, ∞)
b. Using the graph of y = ln x , compress the
graph horizontally by a factor of 1 , and
2
shift up 3 units.
y
b. Using the graph of y = ln x , shift the graph
to the left 3 units and compress vertically by
a factor of 1 .
2
x = −3 y
5
5
(0.5, 3)
(−2, 0)
5 x
(0, 0.55)
−5
−5
c.
Range: (−∞, ∞)
Vertical Asymptote: x = −3
1
ln ( x + 3)
2
1
y = ln ( x + 3)
2
1
x = ln ( y + 3)
2
2 x = ln ( y + 3)
d.
f ( x) =
d.
Inverse
y = e2 x − 3
f
( x) = e
2x
−3
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = e x , compress
Range: (−∞, ∞)
Vertical Asymptote: x = 0
f ( x ) = 3 + ln(2 x)
y = 3 + ln(2 x)
x = 3 + ln(2 y )
x − 3 = ln(2 y )
Inverse
2 y = e x −3
1
y = e x −3
2
1
f −1 ( x) = e x −3
2
y + 3 = e2 x
−1
5 x
−5
−5
c.
(4, 5.08)
horizontally by a factor of 12 , and shift
e.
Range of f: (−∞, ∞)
f.
Using the graph of y = e x , shift to
horizontally to the right 3 units, and
compress vertically by a factor of 12 .
y
5
down 3 units.
y
(5.08, 4)
(3, 0.5)
5
5 x
−5
(0, 0.55)
−5
(0, −2)
5 x
−5
y = −3
−5
700
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
41− 2 x = 2
41.
(2 )
2 1− 2 x
45.
x −3 = 64
=2
2−4 x
(x )
−3 −1/ 3
1
2
=2
2 − 4x = 1
− 4 x = −1
1
x=
4
42.
8
(2 )
3 6+3 x
⎧1 ⎫
⎨ ⎬
⎩4⎭
.
46. log
x=
=
2
(
2 x − x2
48.
)
(
5x + 2 = 7 x −2
)
(
)
( x + 2) ln 5 = ( x − 2) ln 7
x ln 5 + 2 ln 5 = x ln 7 − 2 ln 7
x ln 5 − x ln 7 = − 2 ln 7 − 2 ln 5
x(lng 5 − ln 7) = − 2 ln 7 − 2 ln 5
− 2 ln 7 − 2 ln 5
x=
ln 5 − ln 7
− 2 ( ln 7 + ln 5 )
=
ln 5 − ln 7
2 ( ln 7 + ln 5 )
=
≈ 21.133
ln 7 − ln 5
⎪⎧ 2 ( ln 7 + ln 5 ) ⎪⎫
The solution set is ⎨
⎬ ≈ {21.133} .
⎩⎪ ln 7 − ln 5 ⎭⎪
1
2
) = 2 −1
2x − 2x −1 = 0
− (−2) ± (−2) 2 − 4(2)(−1)
2(2)
2 ± 12 2 ± 2 3 1 ± 3
=
=
4
4
2
The solution is
(
ln 5 x + 2 = ln 7 x − 2
⎧ −1 − 3 −1 + 3 ⎫⎪
,
is ⎪⎨
⎬ ≈ {−1.366, 0.366} .
2 ⎪⎭
⎩⎪ 2
2 x − 2 x 2 = −1
=
( )
x ln 5 = ( x + 2) ln 3
x ln 5 = x ln 3 + 2 ln 3
x ln 5 − x ln 3 = 2 ln 3
x(ln 5 − ln 3) = 2 ln 3
2 ln 3
x=
≈ 4.301
ln 5 − ln 3
⎧ 2 ln 3 ⎫
The solution set is ⎨
⎬ ≈ {4.301} .
⎩ ln 5 − ln 3 ⎭
2
x=
⎧1 ⎫
⎨ ⎬.
⎩8 ⎭
ln 5 x = ln 3x + 2
− 2 ± 12 −2 ± 2 3 −1 ± 3
=
=
4
4
2
2
1
8
5 x = 3x + 2
47.
⎧ 16 ⎫
⎨− ⎬ .
⎩ 9⎭
= 3
4x− x =
.
−6
The solution set is
− 2 ± 22 − 4(2)(−1)
2(2)
44.
⎧1 ⎫
⎨ ⎬
⎩4⎭
x = −6
= 2−3 =
x2 + x
The solution
2
1
4
1/ 2 −6
= 31/ 2
1
x2 + x =
2
2 x2 + 2 x − 1 = 0
3
=
( 2)
= (2 )
= 22
The solution set is
3
64
x=
218+ 9 x = 22
18 + 9 x = 2
9 x = −16
16
x=−
9
43.
1
3
The solution set is
=4
x2 + x
= 64−1/ 3
x=
The solution set is
6+3 x
log x 64 = −3
⎪⎧1 − 3 1 + 3 ⎫⎪
,
⎨
⎬ ≈ {−0.366,1.366} .
2 ⎪⎭
⎪⎩ 2
701
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
49.
92 x = 273 x − 4
(3 )
2 2x
( )
= 33
53.
3x −4
( )
23 = 22
34 x = 39 x −12
4 x = 9 x − 12
−5 x = −12
12
x=
5
23 = 22 x
50.
25
(5 )
2 2x
=5
2
−12
54 x = 5 x
2
−12
54.
51. log3 x − 2 = 2
x − 2 = 32
x−2 =9
x − 2 = 92
x − 2 = 81
x = 83
55. log 6 ( x + 3) + log 6 ( x + 4) = 1
log 6 ( ( x + 3)( x + 4) ) = 1
( x + 3)( x + 4) = 61
Check: log3 83 − 2 = log3 81
= log3 9
=2
The solution set is {83} .
x 2 + 7 x + 12 = 6
x2 + 7 x + 6 = 0
( x + 6)( x + 1) = 0
x = − 6 or x = −1
2 x +1 ⋅ 8− x = 4
Since log 6 (−6 + 3) = log 6 (−3) is undefined, the
solution set is {−1} .
= 22
2 x +1 ⋅ 2−3 x = 22
−2 x +1
2
=2
− 2x +1 = 2
− 2x = 1
)
ln 2 x + ln 5 = ln10 x
x ln 2 + ln 5 = x ln10
ln 5 = x ln10 − x ln 2
ln 5 = x(ln10 − ln 2)
ln 5
=x
ln10 − ln 2
ln 5 ln 5
=
=1
x=
10 ln 5
ln
2
The solution set is {1} .
The solution set is {−2, 6} .
( )
(
2 x ⋅ 5 = 10 x
ln 2 x ⋅ 5 = ln10 x
x 2 − 4 x − 12 = 0
( x − 6)( x + 2) = 0
x = 6 or x = − 2
2 x +1 ⋅ 2
+5 x
⎧ 1⎫
The solution set is ⎨−3, ⎬ .
⎩ 2⎭
x 2 −12
= 5x
3 −x
⋅ 25 x
0 = 2 x2 + 5x − 3
0 = (2 x − 1)( x + 3)
1
or x = −3
x=
2
4 x = x 2 − 12
52.
2
x2
3 = 2 x2 + 5x
⎧12 ⎫
The solution set is ⎨ ⎬ .
⎩5⎭
2x
2
8 = 4 x ⋅ 25 x
56.
2
log(7 x − 12) = 2 log x
log(7 x − 12) = log x 2
7 x − 12 = x 2
x 2 − 7 x + 12 = 0
( x − 4)( x − 3) = 0
x = 4 or x = 3
Since each original logarithm is defined, for
x = 3 and x = 4 , the solution set is {3, 4} .
1
x=−
2
⎧ 1⎫
The solution set is ⎨− ⎬ .
⎩ 2⎭
702
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
57. e1− x = 5
1 − x = ln 5
− x = −1 + ln 5
x = 1 − ln 5 ≈ −0.609
The solution set is {1 − ln 5} ≈ {−0.609} .
58.
61. a.
Using the graph of y = log 2 x , shift the
graph right 2 units and up 1 unit.
y x=2
10
f (x ) = log 2 (x − 2 ) + 1
e1− 2 x = 4
1 − 2 x = ln 4
− 2 x = −1 + ln 4
1 − ln 4
x=
≈ −0.193
2
⎧1 − ln 4 ⎫
The solution set is ⎨
⎬ ≈ {−0.193} .
⎩ 2 ⎭
x
(6, 3)
−10
b.
(3 )
(3 )
+ 4 ⋅ 3x − 3 = 0
x 2
+ 4 ⋅ 3x − 3 = 0
The solution set is {3}. The point ( 6,3) is
on the graph of f.
c.
Let u = 3x .
=
log 2 ( x − 2 ) = 3
x − 2 = 23
x−2 =8
x = 10
The solution set is {10}. The point (10, 4 )
2
−(4) ± (4) − 4(1)(−3)
2(1)
−4 ± 28 −4 ± 2 7
=
= −2 ± 7
2
2
is on the graph of f.
or 3x = −2 + 7
3 x = −2 − 7
3x can't be negative
(
)
7 )} ≈ {−0.398}
d.
x = log3 −2 + 7
{ (
The solution set is log3 −2 +
log 2 ( x − 2 ) = −1
.
x − 2 = 2 −1
1
x−2 =
2
5
x=
2
Based on the graph drawn in part (a),
5
f ( x ) > 0 when x > . The solution set is
2
5
5
⎧
⎫
⎛
⎞
⎨ x | x > ⎬ or ⎜ , ∞ ⎟ .
2⎭
⎝2 ⎠
⎩
60. 4 − 14 ⋅ 4 = 5
Multiply both sides of the equation by 4 x .
(4 )
x 2
− 14 ⋅ 4− x ⋅ 4 x = 5 ⋅ 4 x
(4 )
x 2
(4 ) − 5⋅ 4
( 4 − 7 )( 4
− 14 = 5 ⋅ 4 x
x 2
x
− 14 = 0
x
x
+2 =0
4x − 7 = 0
4x = 7
x = log 4 7
f ( x) = 0
log 2 ( x − 2 ) + 1 = 0
−x
x
f ( x) = 4
log 2 ( x − 2 ) + 1 = 4
u 2 + 4u − 3 = 0
a = 1, b = 4, c = −3
u=
f ( 6 ) = log 2 ( 6 − 2 ) + 1
= log 2 ( 4 ) + 1 = 2 + 1 = 3
x
2 x
(10, 4)
10 x
−10
9 + 4⋅3 − 3 = 0
59.
f ( x ) = log 2 ( x − 2 ) + 1
)
or 4 x + 2 = 0
4 x = −2
4 x can't be negative
The solution set is {log 4 7} ≈ {1.404} .
703
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
e.
f ( x ) = log 2 ( x − 2 ) + 1
log3 ( x + 1) − 4 = 0
y = log 2 ( x − 2 ) + 1
x = log 2 ( y − 2 ) + 1
log 3 ( x + 1) = 4
Inverse
x − 1 = log 2 ( y − 2 )
x + 1 = 34
x + 1 = 81
x = 80
Based on the graph in part (a), f ( x ) < 0 for
y − 2 = 2 x −1
y = 2 x −1 + 2
f −1 ( x ) = 2 x −1 + 2
y x=2
10
−1 < x < 80 . The solution set is
{ x | −1 < x < 80} or ( −1,80 ) .
f −1 ( x) = 2 x −1 + 2
e.
f (x ) = log 2 (x − 2 ) + 1
y=x
62. a.
f ( x ) = log3 ( x + 1) − 4
y = log3 ( x + 1) − 4
y=2
x
10
−10
f ( x) = 0
d.
x = log3 ( y + 1) − 4
Inverse
x + 4 = log3 ( y + 1)
y + 1 = 3x + 4
y = 3x + 4 − 1
−10
f −1 ( x ) = 3x + 4 − 1
f ( x ) = log 3 ( x + 1) − 4
f −1 ( x ) = 3 x + 4 − 1
Using the graph of log 3 x , shift the graph
left 1 unit and down 4 units.
y
y
10
y=x
−10
10
10
x
y = −1
10 x
−10
f (x ) = log 3 ( x + 1) − 4
x = −1
⎛ 760 ⎞
63. h(300) = ( 30(0) + 8000 ) log ⎜
⎟
⎝ 300 ⎠
≈ 3229.5 meters
x = −1
b.
f ( 8 ) = log3 ( 8 + 1) − 4
⎛ 760 ⎞
64. h(500) = ( 30(5) + 8000 ) log ⎜
⎟
⎝ 500 ⎠
≈ 1482 meters
= log3 ( 9 ) − 4 = 2 − 4 = −2
The solution set is {−2}. The point ( 8, −2 )
is on the graph of f.
c.
f (x ) = log 3 ( x + 1) − 4
65. P = 25e0.1d
a. P = 25e0.1(4) = 25e0.4 ≈ 37.3 watts
f ( x ) = −3
log3 ( x + 1) − 4 = −3
b.
log 3 ( x + 1) = 1
50 = 25e0.1d
2 = e0.1d
ln 2 = 0.1d
ln 2
d=
≈ 6.9 decibels
0.1
x + 1 = 31
x +1 = 3
x=2
The solution set is {2}. The point ( 2, −3) is
on the graph of f.
704
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
66. L = 9 + ( 5.1) log d
a.
⎛ r⎞
69. P = A ⎜ 1 + ⎟
⎝ n⎠
L = 9 + ( 5.1) log 3.5 ≈ 11.77
⎛ 0.04 ⎞
= 85, 000 ⎜ 1 +
⎟
2 ⎠
⎝
14 = 9 + ( 5.1) log d
b.
5 = ( 5.1) log d
5
≈ 0.9804
5.1
d ≈ 100.9804 ≈ 9.56 inches
log d =
67. a.
n=
b.
n=
log ( 0.5i ) − log ( i )
log(1 − 0.15)
⎛ 0.5i ⎞
log ⎜
⎟
⎝ i ⎠ = log 0.5 ≈ 4.27 years
=
log 0.85
log 0.85
b.
(2) (18)
71.
≈ $41, 668.97
A = 4000e0.10436(20) ≈ $32, 249.24
The bank’s claim is correct.
A = A0 e kt
0.5 A0 = A0 e k (5600)
0.5 = e5600 k
ln 0.5 = 5600k
ln 0.5
k=
5600
The effective interest rate is computed as
follows:
⎛ 0.04 ⎞
When t = 1, A = 10, 000 ⎜ 1 +
⎟
2 ⎠
⎝
= 10, 000(1.02) 2
= $10, 404
−2(18)
5000 = 620.17e r (20)
5000
≈ e 20 r
620.17
⎛ 5000 ⎞
ln ⎜
⎟ ≈ 20r
⎝ 620.17 ⎠
⎛ 5000 ⎞
ln ⎜
⎟
620.17 ⎠
r≈ ⎝
≈ 0.10436
20
r ≈ 10.436%
70. a.
log10, 000 − log 90, 000
≈ 9.85 years
log(1 − 0.20)
⎛ 0.04 ⎞
68. In 18 years, A = 10, 000 ⎜ 1 +
⎟
2 ⎠
⎝
= 10, 000(1.02)36
≈ $20,398.87
− nt
(2) (1)
0.05 A0 = A0
⎛ ln 0.5 ⎞
⎜
⎟t
e⎝ 5600 ⎠
⎛ ln 0.5 ⎞
⎜
⎟t
2t
0.05 = e⎝ 5600 ⎠
⎛ ln 0.5 ⎞
ln 0.05 = ⎜
⎟t
⎝ 5600 ⎠
ln 0.05
≈ 24, 203
t=
⎛ ln 0.5 ⎞
⎜
⎟
⎝ 5600 ⎠
The man died approximately 24,203 years ago.
(1.02 )2t
72. Using u = T + (u0 − T )e kt , with t = 5 , T = 70 ,
u0 = 450 , and u = 400 .
10, 404 − 10, 000
404
=
= 0.0404 , so
10, 000
10, 000
the effective interest rate is 4.04%.
Note,
In order for the bond to double in value, we have
the equation: A = 2 P .
⎛ 0.04 ⎞
10, 000 ⎜ 1 +
⎟ = 20, 000
2 ⎠
⎝
=2
2t ln1.02 = ln 2
ln 2
t=
≈ 17.5 years
2 ln1.02
400 = 70 + (450 − 70)e k (5)
330 = 380e5 k
330
= e5 k
380
⎛ 330 ⎞
ln ⎜
⎟ = 5k
⎝ 380 ⎠
ln ( 330 / 380 )
k=
5
705
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
Find time for temperature of 150˚F:
150 =
80 =
76. a.
⎛ ln ( 330 / 380 ) ⎞
⎜
⎟t
5
⎠
70 + (450 − 70)e⎝
⎛ ln ( 330 / 380 ) ⎞
⎜
⎟t
5
⎠
380e⎝
0.8
0.8
=
≈ 0.3
1 + 1.67e −0.16 (0) 1 + 1.67
In 2006, about 30% of cars had a GPS.
P (0) =
b. The maximum proportion is the carrying
capacity, c = 0.8 = 80%.
⎛ ln ( 330 / 380 ) ⎞
⎜
⎟t
80
5
⎠
= e⎝
380
⎛ 80 ⎞ ⎛ ln ( 330 / 380 ) ⎞
ln ⎜
⎟t
⎟=⎜
5
⎝ 380 ⎠ ⎝
⎠
c.
0
0.5 = e5.27 k
ln 0.5 = 5.27k
ln 0.5
k=
5.27
In 40 years: A
⎛ r⎞
75. A = P ⎜ 1 + ⎟
⎝ n⎠
)
0.8
= 1 + 1.67e−0.16 t
0.75
0.8
− 1 = 1.67e −0.16 t
0.75
0.8
−1
0.75
= e−0.16 t
1.67
⎛ 0.8
⎞
⎜ 0.75 − 1 ⎟
ln ⎜
⎟ = −0.16t
⎝ 1.67 ⎠
⎛ 0.8
⎞
⎜ 0.75 − 1 ⎟
ln ⎜
⎟
t = ⎝ 1.67 ⎠ ≈ 20.13
−0.16
Note that 2006 + 20.13 = 2026.13 , so 75%
of new cars will have GPS in 2026.
0.5 A0 = A0 e k (5.27)
⎛ ln 0.5 ⎞
⎜
⎟ (40)
= 100e⎝ 5.27 ⎠
30
0
(
A = A0 e kt
⎛ ln 0.5 ⎞
⎜
⎟ (20)
1 + 1.67e −0.16 x
d. Find t such that P (t ) = 0.75 .
0.8
= 0.75
1 + 1.67e −0.16 t
0.8 = 0.75 1 + 1.67e −0.16 t
73. P0 = 6, 451, 058, 790 , k = 0.0115 , and
t = 2015 − 2005 = 10
P = P0 e kt = 6, 451, 058, 790e0.0115 (10)
≈ 7, 237, 271,501 people
In 20 years: A = 100e⎝ 5.27 ⎠
0.8
1
⎛ 80 ⎞
ln ⎜
⎟
⎝ 380 ⎠ ≈ 55.22
t=
ln ( 330 / 380 )
5
The temperature of the skillet will be 150˚F after
approximately 55.22 minutes (or 55 minutes, 13
seconds).
74.
Y1 =
≈ 7.204 grams
≈ 0.519 grams
nt
170
77. a.
1(10)
⎛ 0.0425 ⎞
A = 319 ⎜ 1 +
⎟
1 ⎠
⎝
0
= 319 (1.0425 ) ≈ 483.67
10
The government would have to pay back
approximately $483.67 billion in 2015.
150
16
b. Using EXPonential REGression on the data
yields: y = (165.73)( 0.9951)
x
706
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Review Exercises
c.
Y1 = (165.73)( 0.9951)
b. Using LOGISTIC REGression on the data
yields:
46.93
C=
1 + 21.273e −0.7306t
x
170
c.
0
16
150
Y1 =
46.93
1 + 21.273e −0.7306t
50
d. Find x when y = 110 .
(165.73)( 0.9951) x = 110
110
165.73
⎛ 110 ⎞
x ln 0.9951 = ln ⎜
⎟
⎝ 165.73 ⎠
⎛ 110 ⎞
ln ⎜
⎟
165.73 ⎠
x= ⎝
≈ 83
ln 0.9951
Therefore, it will take approximately 83
seconds for the probe to reach a temperature
of 110˚F.
( 0.9951) x =
−1
In reality, all 50 people living in the town
might catch the cold.
e.
40
0
Y1 = 18.9028 − 7.0963ln x
10
40
0
−10
d. If x = 23, then
y = 18.9028 − 7.0963ln 23 ≈ −3o F .
50
−1
0
)
46.9292
= 1 + 21.2733e −0.7306t
10
46.9292
− 1 = 21.2733e −0.7306t
10
3.69292 = 21.2733e −0.7306t
3.69292
= e −0.7306t
21.2733
⎛ 3.69292 ⎞
ln ⎜
⎟ = −0.7306t
⎝ 21.2733 ⎠
⎛ 3.69292 ⎞
ln ⎜
⎟
⎝ 21.2733 ⎠ = t
−0.7306
t ≈ 2.4
Therefore, after approximately 2.4 days
(during the 10th hour on the 3rd day), 10 people
had caught the cold.
b. Using LnREGression on the data
yields: y = 18.9028 − 7.0963ln x where y =
wind chill and x = wind speed.
79. a.
Find t when C = 10 .
46.9292
= 10
1 + 21.2733e −0.7306t
46.9292 = 10 1 + 21.2733e −0.7306t
(
−10
c.
9
d. As t → ∞ , 21.2733e−0.7306t → 0 , which
means 1 + 21.2733e−0.7306t → 1 , so
46.9292
C=
→ 46.9292
1 + 21.2733e −0.7306t
Therefore, according to the function, a
maximum of about 47 people can catch the
cold.
10
78. a.
0
9
The data appear to have a logistic relation
707
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
Find t when C = 46 .
46.9292
= 46
1 + 21.2733e −0.7306t
46.9292 = 46 1 + 21.2733e −0.7306t
f.
(
7. ln133 ≈ 4.890
8.
)
f ( x) = 4 x +1 − 2
a.
46.9292
= 1 + 21.2733e −0.7306t
46
46.9292
− 1 = 21.2733e −0.7306t
46
0.0202 = 21.2733e −0.7306t
0.0202
= e −0.7306t
21.2733
0.0202
= e −0.7306t
21.2733
⎛ 0.0202 ⎞
ln ⎜
⎟ = −0.7306t
⎝ 21.2733 ⎠
⎛ 0.0202 ⎞
ln ⎜
⎟
⎝ 21.2733 ⎠ = t
−0.7306
t ≈ 9.5
Therefore, after approximately 9.5 days
(during the 12th hour on the 10th day), 46
people had caught the cold.
Domain: (−∞, ∞)
b. Using the graph of y = 4 x , shift the graph 1
unit to the left, and shift 2 units down.
(−1, −1)
c.
d.
Range: (−2, ∞)
Horizontal Asymptote: y = −2
f ( x ) = 4 x +1 − 2
y = 4 x +1 − 2
x = 4 y +1 − 2
Inverse
y +1
x+2= 4
y + 1 = log 4 ( x + 2)
y = log 4 ( x + 2) − 1
Chapter 7 Test
1. 3x = 243
x
f −1 ( x) = log 4 ( x + 2) − 1
5
3 =3
x=5
2. logb 16 = 2
b 2 = 16
e.
Range of f: (−2, ∞)
f.
Using the graph of y = log 4 x , shift the
graph 2 units to the left, and shift down 1
unit.
b = ± 16 = ±4
Since the base of a logarithm must be positive,
the only viable solution is b = 4 .
3. log5 x = 4
x = 54
x = 625
4. e3 + 2 ≈ 22.086
5. log 20 ≈ 1.301
6. log3 21 =
ln 21
≈ 2.771
ln 3
708
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Test
9.
f ( x) = 1 − log 5 ( x − 2 )
a.
10. 5 x + 2 = 125
5 x + 2 = 53
x+2=3
x =1
The solution set is {1} .
Domain: (2, ∞)
b. Using the graph of y = log5 x , shift the
graph to the right 2 units, reflect vertically
about the y-axis, and shift up 1 unit.
11. log( x + 9) = 2
x + 9 = 102
x + 9 = 100
x = 91
The solution set is {91} .
12. 8 − 2e − x = 4
−2e − x = −4
c.
Range: (−∞, ∞)
Vertical Asymptote: x = 2
d.
f ( x ) = 1 − log5 ( x − 2 )
e− x = 2
− x = ln 2
x = − ln 2 ≈ −0.693
The solution set is {− ln 2} ≈ {−0.693} .
y = 1 − log5 ( x − 2 )
x = 1 − log5 ( y − 2 )
(
Inverse
2
x +3= x+6
x − 1 = − log 5 ( y − 2 )
2
x − x−3 = 0
1 − x = log5 ( y − 2 )
y − 2 = 51− x
x=
y = 51− x + 2
f.
−(−1) ± (−1) 2 − 4(1)(−3) 1 ± 13
=
2(1)
2
⎪⎧1 − 13 1 + 13 ⎪⎫
,
The solution set is ⎨
⎬
2 ⎭⎪
⎩⎪ 2
f −1 ( x ) = 51− x + 2
e.
)
13. log x 2 + 3 = log ( x + 6 )
Range of f: (−∞, ∞)
≈ {−1.303, 2.303} .
x
Using the graph of y = 5 , reflect the graph
horizontally about the y-axis, shift to the
right 1 unit, and shift up 2 units.
14.
7 x +3 = e x
ln 7 x + 3 = ln e x
( x + 3) ln 7 = x
x ln 7 + 3ln 7 = x
x ln 7 − x = −3ln 7
x(ln 7 − 1) = −3ln 7
−3ln 7
3ln 7
x=
=
≈ −6.172
ln 7 − 1 1 − ln 7
⎧ 3ln 7 ⎫
The solution set is ⎨
⎬ ≈ {−6.172} .
⎩1 − ln 7 ⎭
709
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
15. log 2 ( x − 4 ) + log 2 ( x + 4 ) = 3
18. a.
log 2 ⎡⎣( x − 4 )( x + 4 ) ⎤⎦ = 3
(
)
x 2 − 16 = 23
⎛ 0.05 ⎞
So, A = 1000 ⎜ 1 +
⎟
⎝
12 ⎠
2
x − 16 = 8
x 2 = 24
(12) (2 / 3)
8
Because x = −2 6 results in a negative
arguments for the original logarithms, the only
viable solution is x = 2 6 . That is, the solution
b. Note that 9 months =
3
year. Thus,
4
A = 1000 , r = 0.05 , n = 4 , and t =
{ }
set is 2 6 ≈ {4.899} .
3
. So,
4
(4) (3/ 4)
⎛ 0.05 ⎞
1000 = A0 ⎜ 1 +
⎟
4 ⎠
⎝
1000 = A0 (1.0125)3
1000
≈ $963.42
A0 =
(1.0125)3
⎛
⎞
4 x3
log 2 ⎜⎜ 2
⎟⎟
⎝ x − 3x − 18 ⎠
⎛
⎞
22 x3
= log 2 ⎜⎜
⎟⎟
⎝ ( x + 3)( x − 6) ⎠
)
= log 2 22 x3 − log 2 [ ( x − 6)( x + 3) ]
c.
r = 0.06 and n = 1 . So,
⎛ 0.06 ⎞
2 A0 = A0 ⎜ 1 +
⎟
1 ⎠
⎝
2 A0 = A0 (1.06)t
= log 2 22 + log 2 x3 − [ log 2 ( x − 6) + log 2 ( x + 3)]
= 2 + 3log 2 x − log 2 ( x − 6) − log 2 ( x + 3)
17.
2
.
3
⎛ 0.05 ⎞
= 1000 ⎜ 1 +
⎟ ≈ $1033.82
⎝
12 ⎠
x = ± 24 = ±2 6
(
2
year. Thus,
3
P = 1000 , r = 0.05 , n = 12 , and t =
log 2 x 2 − 16 = 3
16.
Note that 8 months =
(1)t
2 = (1.06)t
A = A0 ekt
ln 2
≈ 11.9
ln1.06
It will take about 11.9 years to double your
money under these conditions.
34 = 50ek (30)
t = log1.06 2 =
30 k
0.68 = e
ln 0.68 = 30k
ln 0.68
k=
30
⎛ ln 0.68 ⎞
⎜
⎟t
50e⎝ 30 ⎠
Thus, the decay model is A =
We need to find t when A = 2 :
19. a.
.
⎛ ln 0.68 ⎞
⎜
⎟t
30 ⎠
2 = 50e⎝
⎛ ln 0.68 ⎞
⎜
⎟t
0.04 = e⎝ 30 ⎠
⎛ ln 0.68 ⎞
ln 0.04 = ⎜
⎟t
⎝ 30 ⎠
ln 0.04
≈ 250.39
t=
⎛ ln 0.68 ⎞
⎜
⎟
⎝ 30 ⎠
There will be 2 mg of the substance remaining
after about 250.39 days.
⎛ I ⎞
80 = 10 log ⎜ −12 ⎟
⎝ 10 ⎠
⎛ I ⎞
8 = log ⎜ −12 ⎟
⎝ 10 ⎠
8 = log I − log10−12
8 = log I − (−12)
8 = log I + 12
−4 = log I
I = 10−4 = 0.0001
If one person shouts, the intensity is 10−4
watts per square meter. Thus, if two people
shout at the same time, the intensity will be
2 × 10−4 watts per square meter. Thus, the
loudness will be
⎛ 2 × 10−4 ⎞
D = 10 log ⎜
= 10 log ( 2 × 108 ) ≈ 83
−12 ⎟
⎝ 10
⎠
decibels
710
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Cumulative Review
5. 2 x − 4 y = 16
b. Let n represent the number of people who
must shout. Then the intensity will be
n × 10−4 . If D = 125 , then
⎛ n × 10−4 ⎞
125 = 10 log ⎜⎜
⎟
−12 ⎟
⎝ 10
⎠
(
125 = 10 log n × 108
(
12.5 = log n × 108
8
)
x-intercept:
2 x − 4 ( 0 ) = 16
2 x = 16
x =8
)
y-intercept:
2 ( 0 ) − 4 y = 16
−4 y = 16
y = −4
12.5
n × 10 = 10
n = 104.5 ≈ 31, 623
About 31,623 people would have to shout at
the same time in order for the resulting
sound level to meet the pain threshold.
6.
Chapter 7 Cumulative Review
Using the graph of y = x3 , shift the graph 1 unit
to the left, stretch vertically by a factor of 3, and
shift 2 units down.
1. The graph represents a function since it passes
the Vertical Line Test.
The function is not a one-to-one function since
the graph fails the Horizontal Line Test.
2.
f ( x) = 3( x + 1)3 − 2
f ( x ) = 2 x 2 − 3x + 1
a.
f ( 3) = 2 ( 3) − 3 ( 3) + 1 = 18 − 9 + 1 = 10
b.
f ( − x ) = 2 ( − x ) − 3 ( − x ) + 1 = 2 x 2 + 3x + 1
c.
f ( x + h) = 2 ( x + h) − 3( x + h) + 1
2
2
2
(
)
= 2 x 2 + 2 xh + h 2 − 3 x − 3h + 1
2
7. a.
2
= 2 x + 4 xh + 2h − 3 x − 3h + 1
g ( x ) = 3x + 2
Using the graph of y = 3x , shift up 2 units.
3. x 2 + y 2 = 1
2
a.
2
1 1 1
⎛1⎞ ⎛1⎞
⎛1 1⎞
⎜ ⎟ + ⎜ ⎟ = + = ≠ 1 ; ⎜ , ⎟ is
4 4 2
⎝2⎠ ⎝2⎠
⎝2 2⎠
not on the graph.
2
2
1 3
⎛1⎞ ⎛ 3⎞
b. ⎜ ⎟ + ⎜⎜
⎟⎟ = + = 1 ;
4 4
⎝2⎠ ⎝ 2 ⎠
the graph.
⎛1 3⎞
⎜⎜ ,
⎟⎟ is on
⎝2 2 ⎠
4. 3 ( x − 2 ) = 4 ( x + 5 )
3 x − 6 = 4 x + 20
−26 = x
The solution set is {−26} .
Domain of g: (−∞, ∞)
Range of g: (2, ∞)
Horizontal Asymptote for g: y = 2
711
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
g ( x ) = 3x + 2
b.
10. a.
y =3 +2
x = 3y + 2
log3 ( x + 2 ) = 0
x + 2 = 30
x + 2 =1
x = −1
The solution set is {−1} .
x
Inverse
y
x−2 =3
y = log3 ( x − 2)
g −1 ( x ) = log3 ( x − 2 )
b.
log3 ( x + 2 ) > 0
x + 2 > 30
x + 2 >1
x > −1
Domain of g −1 : (2, ∞)
Range of g −1 : (−∞, ∞)
Vertical Asymptote for g −1 : x = 2
The solution set is { x x > −1} .
c.
c.
log3 ( x + 2 ) = 3
x + 2 = 33
x + 2 = 27
x = 25
The solution set is {25} .
11. a.
8.
20
4 x − 3 = 82 x
(2 )
2 x −3
( )
= 23
2x
0
2 2 x − 6 = 26 x
2x − 6 = 6x
−6 = 4 x
6
3
x=− =−
4
2
⎧ 3⎫
The solution set is ⎨− ⎬ .
⎩ 2⎭
0
80
b – c. Answers will vary.
Chapter 7 Projects
Project I
a. Newton’s Law of Cooling:
u (t ) = T + (u0 − T )e kt , k < 0
Container 1: u0 = 200ºF, T = 70ºF, u(30)=100ºF,
t = 30 mins.
100 = 70 + (200 − 70)e30 k
9. log3 ( x + 1) + log3 (2 x − 3) = log 9 9
log3 ( ( x + 1)(2 x − 3) ) = 1
( x + 1)(2 x − 3) = 31
2 x2 − x − 3 = 3
30 = 130e30 k
30
= e30 k
130
⎛ 30 ⎞
30k = ln ⎜
⎟
⎝ 130 ⎠
1 ⎛ 30 ⎞
k = ln ⎜
⎟ ≈ −0.04888
30 ⎝ 130 ⎠
2
2x − x − 6 = 0
( 2 x + 3) ( x − 2) = 0
3
or x = 2
2
⎛ 3 ⎞
⎛ 1⎞
Since log3 ⎜ − + 1⎟ = log3 ⎜ − ⎟ is undefined
⎝ 2 ⎠
⎝ 2⎠
the solution set is {2} .
x=−
u1 (t ) = 70 + 130e −0.04888t
712
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7 Projects
Container 2:
⎛ 130 − 60 ⎞
ln ⎜
⎟
200 − 60 ⎠
≈ 16.83 minutes
t= ⎝
−0.04118
Container 2: u0 = 200ºF, T = 60ºF, u(25)=110ºF,
t = 25 mins.
100 = 60 + (200 − 60)e25k
50 = 140e25k
50
= e25k
140
⎛ 50 ⎞
25k = ln ⎜
⎟
⎝ 140 ⎠
⎛ 50 ⎞
ln ⎜
⎟
140 ⎠
k= ⎝
≈ −0.04118
25
u2 (t ) = 60 + 140e−0.04118t
Container 3:
⎛ 130 − 65 ⎞
ln ⎜
⎟
200 − 65 ⎠
t= ⎝
≈ 16.28 minutes
−0.04490
c. Container 1:
⎛ 110 − 70 ⎞
ln ⎜
⎟
130 − 70 ⎠
t= ⎝
≈ 8.295
−0.04888
It will remain between 110º and 130º for about
8.3 minutes.
Container 3: u0 = 200ºF, T = 65ºF, u(20)=120ºF,
t = 20 mins.
100 = 65 + (200 − 65)e 20 k
Container 2:
⎛ 110 − 60 ⎞
ln ⎜
⎟
130 − 60 ⎠
t= ⎝
≈ 8.171
−0.04118
It will remain between 110º and 130º for about
8.17 minutes
55 = 135e20 k
55
= e20 k
135
⎛ 55 ⎞
20k = ln ⎜
⎟
⎝ 135 ⎠
⎛ 55 ⎞
ln ⎜
⎟
135 ⎠
k= ⎝
≈ −0.04490
20
u3 (t ) = 65 + 135e −0.04490t
Container 3:
⎛ 110 − 65 ⎞
ln ⎜
⎟
130 − 65 ⎠
t= ⎝
≈ 8.190
−0.04490
It will remain between 110º and 130º for about
8.19 minutes.
b. We need time for each of the problems, so solve
for t first then substitute the specific values for
each container:
u = T + (u0 − T )e kt
d. All three graphs basically lie on top of each
other.
u − T = (u0 − T )ekt
u −T
= ekt
u0 − T
e. Container 1 would be the best. It cools off the
quickest but it stays in a warm beverage range
the longest.
⎛ u −T ⎞
kt = ln ⎜
⎟
⎝ u0 − T ⎠
⎛ u −T ⎞
ln ⎜
⎟
u −T ⎠
t= ⎝ 0
k
f. Since all three containers are within seconds of
each other in cooling and staying warm, the cost
would have an effect. The cheaper one would be
the best recommendation.
Container 1:
⎛ 130 − 70 ⎞
ln ⎜
⎟
200 − 70 ⎠
t= ⎝
≈ 15.82 minutes
−0.04888
713
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 7: Exponential and Logarithmic Functions
Project II
Solder Joint
Strain, εp
0.01
0.035
0.1
0.4
X=ln(εp) Fatigue
Cycles, Nf
10, 000
−4.605
1000
−3.352
100
−2.303
10
−0.916
9.210
6.908
4.605
2.303
1.5
0.405
0
1
6. Nf = e0.63 (0.02) −1.84
Nf = 2510.21 cycles
Y=ln(Nf)
Nf = e0.63 (ε p ) −1.84
3000 = e0.63 (ε p) −1.84
3000
= (ε p ) −1.84
e0.63
⎛ 3000 ⎞
⎟
⎝ e0.63 ⎠
ε p = 0.018
εp=⎜
−1
1.84
11,000
1.
7. Nf = e0.63 (ε p ) −1.84
−0.1
Nf = 1.88(ε p ) −1.84
Nf
= (ε p ) −1.84
1.88
1.6
−1000
11,000
2.
ε p = ( 0.53Nf )
−1
1.84
ε p = ( 0.53Nf )
−.543
ε p = 1.41( Nf )
−5
ε p = 1.41( Nf )
−.543
ε p = 1.41( 3000 )
ε p = 0.018
−.543
−.543
1
−1000
Project III
The shape becomes exponential.
Chart: Answers will vary depending upon the values for
the car chosen.
10
3.
a. Answers will vary.
−5
b. Answers will vary, but in general, they will be in
the form y = ab x
1
−1
c. A = A0 bt
The shape became linear.
A = A0 ert
10
4.
−5
∴ er = b
r = ln b
A = A0 e(ln b )t
1
d. A0 is the purchase price of the vehicle, or a close
approximation to it.
−1
−1
Y = −1.84 X + 0.63
e. Answers will vary.
5. Y = −1.84 X + 0.63
ln( Nf ) = −1.84 ln(ε p ) + 0.63
(
)
f. Answers will vary.
g. Answers will vary. However, one might consider
how well a car will “hold its value” over time. In
other words, one would prefer to purchase a
vehicle with a low depreciation rate.
ln( Nf ) = ln (ε p ) −1.84 + ln(e0.63 )
((
)
ln( Nf ) = ln (ε p) −1.84 (e0.63 )
(
Nf = (ε p)
−1.84
) (e
0.63
)
)
Nf = e0.63 (ε p ) −1.84
714
© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.