160
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
Section 3.4 Solving Exponential and Logarithmic Equations
[] To solve an exponential equation, isolate the exponential expression, then take the logarithm of both sides. Then
solve for the variable.
1. loga ax = x
2. lne~ = x
[] To solve a logarithmic equation, rewrite it in exponential form. Then solve for the variable.
1. al°g~x -" x
2. elnx -- x
[] If a > 0 and a 4:1 we have the following:
1. loga x = loga y ==~ x ~ y
2. aX=ar =~ x=y
[] Use your graphing utility to approximate solutions.
,,
Solutions to Odd-Numbered Exercises
1, 4~-7 = 64
(a) x = 5
42<5)-7 = 43 = 64
Yes, x = 5 is a solution.
3. 3ex+2 = 75
(a) x=-2+e25
3e(-Z+e2~)+~ =., 3ee2~ 4:~75
No, x = -2 + e~5 is not a solution.
(b) x=-2+In25
3e(-z+m25)+2 = 3et~25 = 3(25) = 75
Yes, x = -2 + In 25 is a solution.
(b) x = 2
1
42(2)-7 = 4-3 = 6-~ ~:
64
No, x = 2 is not a solution.
5. log4(3x) = 3 ==~ 3x = 43 ~ 3x = 64
(a) x ~ 20.3560
3(20.3560) = 61.0680 4:64
No, x ~ 20.3560 is not a solution.
(b) x = -4
3(-4) = - 12 :~ 64
No, x = -4 is not a solution.
(e) x ~ 1.2189
3e1’2189+2 = 3e3"2189 -~ 75
Yes, x ~ 1.2189 is a solution.
7. ln(x- 1)= 3.8
(a) x= 1 +e3’8
ln(1 + e3’8 - 1) = In e3’8 = 3,8
Yes, x = 1 + e3’8 is a solution.
(b) x ~- 45.7012
1n(45.7012- 1) = ln(44.7012)~- 3.8
Yes, x ~- 45.7012 is a solution.
3(-~) = 64
Yes, x = ~ is a solution.
(c) x= 1+1n3.8
ln(1 + In 3.8 - 1) = In(In 3.8) ~ 0.289
No, x = 1 + In 3.8 is not a solution.
161
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
11,
11
o
g
-1
Intersection Point: (3, 8)
Algebraically, 2x = 8
14
Intersection Point: (9, 2)
Algebraically, log3 x = 2
2x=23
x = 32 = 9==~y = 2=0(9,2)
x= 3==>y = 8:=0(3,8)
15. 4:’ = 16
4x= 42
x=2
Intersection Point: (5, 0)
Algebraically, In (x - 4) = 0
x-4=e°= 1
x = 5 =0y 0=0(5,0)
17. 5x=625
5x -- 54
19. 8x=4
3x-1 = 33
8x --- 82/3
x-1=3
x=4
x=~2
x=4
25. lnx-ln5=O
In x = In 5
29. lnx =-7
27, e"=4
x = In 4 ~- 1.386
x=e-7
31. logx 625 = 4
:d = 625
X4 = 54
x=5
x=5
33. loglo x = - 1
x = 10-1
1
x -- 10
35. ln(2x- 1)= 0
eO=2x- 1
1=2x-1
2=2x
l=x
37. lne~=x21ne=x2
39. el~(5x+2) = 5x + 2
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
43.
45. e~ = 10
x = In 10 ~ 2.303
10~ = 570
loglo llY = loglo 570
x = lOglo 570 ~- 2.756
1
47. 5-t/2 = 0.20 =5-
49. 23-x = 565
(3 - x) In 2 = In 565
-xln2 = In 565 - 3 In2
x = (31n 2 - In 565)/ln 2 ~- - 6.142
_tin5 = -In5
2
t
t=2
53. 7 - 2ex = 5
-2ex = -2
eX=l
x=lnl=O
51. 500e-x = 300
-x = In-35
x = -ln]= ln~ ~ 0.511
55.
ez~- 4ex- 5 = 0
(e~-5)(e~+ 1)=0
ex = 5 or ex = -1
x = In 5 ~- 1.609
(e~ = - 1 is impossible.)
57. 50(120 - ex/~) = 600
120 - ex/2 = 12
ex/2 = 108
- = In 108
2
x = 2In 108 ~ 9.364
59. Using the root feature of a graphing utility for
y= 1+ 12/0.10/~t
-2=0,
you obtain t ~ 6.960.
61.
f(x)
0.6
0.7
6.05
8.17
0.8
0.9
1.0
63.
5
1756 1598
11.02 14.88 20.09
600
o
x ~ 0.828
6
x ~ 8.635
7
8
1338
908
9
200
163
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
67. 2-3x = 0.90
65. 23X = 50
¯
Graphing y = 2-3x - 0.90, you obtain x ~ 0.051
Graphing y - 23x - 50, you obtain x ~- 1.881
69. 5(10~-6) = 7
365 ] =4==~t=21.330
Graphing y = 5(10~-6) - 7, you obtain x ~- 6.146 71. 1 + 0.065~365t
3000
73. 2+e
- z~
2
77. g (t) = eO’09t - 3
75. g(x) = 6el-X- 25
1500 = 2 + ez~
1498 = ez~
In 1498 = In ez~
In 1498 = 2x
!-
!i
!ii
In 1498
- x ~ 3.656
2
zero at x = -0.427
79° lnx =-3
x=e-3~-0.050
zero at t = 12.207
81. In 4x = 2.1
4x = e2’1
¯ X -" ¼e2"1
~- 2.042
83. 21n 3x = 19
19
ln3x = w = 9.5
2
85. lOglo(Z - 3) = 2
Z-3=102
Z= 10~+3 = 103
3x = e9’5
x = ½e9"5 ~ 44531242
87. 7 log4 (0.6x) = 12
12
loga(O.6x) = T
3
412/7 = 0.6x = -=x
89. lnv/~ + 2 = 1
.v/~+ 2 = el
x+2=e~
x = e~ - 2 ~ 5.389
5 412/7 ~. 17.945
x=~
91. ln(x + 1)z = 2
eln(x+l)2 -. e2
(x+ 1)2=e~
x+ 1 =eorx+ 1 =-e
x=e- 1~1.718
o1"
x=-e- 1~--3.718
x = 2(x- 1)
x=2x- 2
164
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
ln(x + 5) = ln(x- 1) - ln(x + 1)
95.
ln(x + 5) = in(x- 1I
x+5-
X-1
x+l
(x+5)(x+ 1)= x-1
x2 + 6x + 5 =x-1
x2 + 5x + 6 =0
(x+2)(x+3) = 0
x= -2 orx= -3
Both of these solutions are extraneous, so the equation has no solution.
97. loglo 8x- loglo(1 + -,/’~) = 2
8x
l°gml + w/~=2
8x
1 +,f~- 102
8x = 100 + 100,,/~
8x- 100,/~- 100 = 0
25./ - 25 = o
25 +_ ~/252 - 4(2)(-25)
_._
25 _+5./55
4
4
Choosing the positive value, we have v/~ ~ 13.431 and x ~ 180.384.
99.
2
1.39
x
f(x)
3
1.79
4
2.08
5
2.30
6
2.49
101.
X
12
f(x)
9.79
13
14
15
10.22 10.63 11.00
16
11.36
"14
x ~ 5.512
iii
-2
o
o
2~
105. 31nx=5
103. loglo(Z- 4) = 1
Graphing y = 3 In x - 5, you obtain x ~-- 5.294.
Graphing y = loglo(z - 4) - 1, you obtain z = 14.
107. lnx+ln(x-3)= 1
Graphing y = In x + ln(x - 3) - !, you obtain
x ~ 3.729.
109. ln(x- 5)= ln(x- 3) - ln(x + 3)
Graphing y = ln(x- 5) - ln(x- 3) + ln(x + 3),
you obtain x ~ 5.275.
165
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
113. yl = 8
111. Yl = 7
Y2 "- 2x
= 4e-°’2x
-9
From the graph we have (x, y) ~ (2.807, 7).
From the graph, we have (x, y) ~ (- 3.466, 8).
115. Yl = 3
y2= lnx
I
117. (a)
||
!
(b) 3000 = lO00e°’°85t
A = Pert
2000 = lO00e°’°~
2 = e0"085t
In 2 = 0.085t
In 2
-t
0.085
t = 8.2 years
3 = e0"085t
In 3 = 0.085t
In 3
0.085
~ "" t
t ~ 12.9 years
,
From the graph we have
(x, y) ~ (20.086, 3).
(b) From the graph we see horizontal asymptotes at
y = 0 and y = 100. These represent the lower
and upper percent bounds.
119. (a) .o
0
110
o
Females:
(c) Males:
50=
100
1 + e-0’6114(x-69"71) = 2
e-0"6114(x-69"71) = 1
-0.6114(x - 69.71) = In 1
-0.6114(x - 69.71) = 0
x = 69.71 inches
121. p = 500 - 0.5(e°’°°4x)
(a) p = 350
350 = 500 - 0.5(e°’°°4x)
300 = e°’°°4x
O.O04x = In 300
x ~ 1426 units
50 =
1 + e-0"6114(x-69’71)
100
1 + e-0’666°7(x-64"51)
1 + e-0’66607(x-64’51) =
2
-°’666°7(x-64"51)
e
=1
-0.66607(x - 64.51) = In 1
-0.66607(x - 64.51) = 0
x = 64.51 inches
(b)
p = 300
300 = 500 - 0.5(e°’°°40
400 = e°’°°4x
O.O04x = In 400
x ~- 1498 units
166
PART I: Solutions to Odd-Numbered Exercises and Practice Tests
123. V = 6.7e-4s’l/t, t > 0
125. T = 2011 + 7(2-h)]
(a) "°
(a)
o
1,~oo
o
o
(b) As t -----> oo, V---> 6.7.
o
(b) We see a horizontal asymptote at y = 20.
This represents the room temperature.
Horizontal asymptote: y = 6.7
The yield will approach
6.7 million cubic feet per acre.
(c) 1.3 = 6.7e-4s’l/t
1.3
6.7
(c)
100 = 2011 + 7(2-h)]
5 = 1 + 7(2-h)
4 = 7(2-h)
4
-48.1
t = ~ ~ 29.3 years
In (1~-~)
ln(4/7)
-h
-In 2
h ~ 0.81 hour
(c) y = e2’7061nx-1’175 "- e~’7°~-1"175 = e-l’175x2"706
127. (a)
--- 0.309X2.706
The second
model is better.
2
0
y = 15.17x - 46.15
8
y = 100 when x ~ 9.6, or during 1999.
(b)
lnx
1.0986
lny
2.0541 2.2300
5,0
1.3863
1.6094 1.7918
1.9459
3.0681 3.7658 4.2002
In y = 2.706 In x - 1.175
For y = 100, In 100 = 2.706 In x - 1.175
5.780 = 2.706 In x
In x = 2.136
x ~ 8.5 or during 1998
167
PART 1: Solutions to Odd-Numbered Exercises and Practice Tests
129. False. A logarithmic equation
can have any number of
extraneous solutions
131. To find the length of time it
o takes for an investment P to
double to 2PI solve
2P = Pert
2=en
133. f(x) = -3-x-3 + 5
9
8
7
6.
ln2 = rt
In 2
m t.
r
Thus, you can see that the
time is not dependent on the
size of the investment, but
rather the interest rate.
In 22
137. log322 = ln3 ~ 2.814
Section 3.5 Exponential and Logarithmic Models
[] You should be able to solve compound interest problems.
1. A=PI+
2. A = Per~
[] You should be able to solve growth and decay problems.
(a) Exponential growth if b > 0 and y = ae~’x.
(b) Exponential decay if b > 0 and y = ae-~’x.
[] You should be able m use the Gaussian m~el
y = ae-(~-~)Vc.
You should be able to use the logistics ~owth model
a
Y = 1 + be-(x-~)/a"
You should be able to use the logarithmic models
y = ln(~ + b) and y = log~o(~ + b).
I1~
12345
140
139. log21 140 = InIn2-----i-=
1.623