chapter
53.
48 ,-" 0
+3) , 0
3
log, ( x + 5) = 3
43=x+5
59 = x
8.
+3..0
0 01 el
2x __
54. log,(x -7) = 2
3
=111(-3)
In e"
ln(-3) does not exist.
6
6
52 = x -7
32 = x
55. log,(x - 4) = -3
z 090
2
+3x-2=0
1)(3x -
I)
=°
3' _1=0
=0
3' = l
=log]
3 3 .,.• -2
= log( 2)
lox
xlog3 = 0
t do
X
=
X
=
0
log3
0
3 -3 = x -4
1 = x -4
27
1
4— = x
27
4.04 = x
56. l og , (x + 2) = 7 -2 = x + 2
=x+2
49
, 48
-I— = x
49
-1.98= x
1
he solution set is {0}.
22'
+2'
-12 = 0
(2' + 4)(2 2 -3) = 0
2x+4=0
2' = -4
1112' =111(-4)
2' -3 = 0
2=
57. log,(3x + 2) = 3
4'=3x+2
64 =3x+2
62 3x
62 re x
3
20.67 = x
ln 2' =11)3
xln 2 = 1n3
can't do
x = 1113
In 2
x =1.58
58. log,(4x +1) = 5
2.2=4x+1
32=4x+1
40. log, x = 4
34 = x
31= 4x
8 1= x
31_
4 -x
7.75 = x
50. log, x = 3
5' = x
125= x
59. 51n 2x = 20
1n2x=4
51. In x = 2
e 2 .__ x
7.39 =
2
e("x = e4
2x = e4
x
x = ?- =27.30
52. lnx=3
=x
20.09 = x
447
, ©2 007
is protected under all copyright laws as they currently exist.
rights reserved. This materia l
NJ. AU
portion of this material may be reproduced,
in any form or by any means, without permission in writing from the publisher.
Pearson Education, Inc., Upper Saddle River,
No
ISM: Preecelag
Chapter 3
66. log, (x+5)+ log, x = 2
log, x(x + 5) = 2
x(x + 5) = 62
x 2 +Sx=36
x2 +5x-36=0
(x + 9)(x — 4) = 0
x = -9 or x = 4
x = —9 does not check because
log,(-9+ 5) does not exist.
The solution set is {4}.
60. 61t1 2x = 30
In 2x = 5
5
en2x
=e
2x = es
x=
61.
2
74.21
6+2Inx=5
21n x = —1
1
Inn -2
67. log, (x —5)+ log, (x +3) = 2
elni =eve
x=
XX
log, Rx — 5)(x + 3)] = 2
0.61
(x — 5)(x +3) =32
x2 —2x-15=9
x2-2x-24=0
— 6)(x + 4) = 0
x = 6 or x = —4
x = -4 does not check because log, (-4 —5)
does not exist. The solutionet is {6}.
62. 7+3lnx=6
31n x = —1
lnx = --1
3
el" =e yi
0.72
x
68. log, (x —0+ log, (x + 1) = 3
63. In x+3=1
e ln3,1c . 3 = eI
E. 3
log, Rx —1)(x +In=3
=e
(x —1)(x +1) =
x+3=e2
x = e2 —3 A% 4.39
64. In
x2 —1= 8
x2 = 9
x=3orx=-3
x = —3 does not check because
loge(-3 —1) does not exist.
The solution set is {3}.
4 =1
e ln“17+71 = e l
4=e
x +4 = e2
x=e2-43.39.
69. log e (x + 2)— log2 (x —5) = 3
loge H1 — 3
x-5
x +2
65. log, x + log, (4x —1) =1
x-5
x+2
log, {4x2 — x) =1
4x2 —x=5
4x2 —x — 5 = 0
(4x — 5)(x + 1) = 0
5
x=- orx=-1
4
x = —1 does not check because log, (-1) does
not exist.
{5
The solution set is —}.
4
2'
—8
x-5
x + 2 = 8(x —5)
x+2=8x-40
7x=42
x=6
448
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3
ISM: Precalculus
70.
log,
x+ 2
1°g°
1
+ 2)— log, (x —
x —1
72.
3 log,(x —1) = 5— 2
=1
3 log,(x —1) = 3
x+2
log 2 (x —1) =1
x —1
x+2
3 log, (x —1) = 5 — log, 4
2' = x — I
3=x
=4
x —1
x+2=4(x-1)
x+ 2 = 4x — 4
3x = 6
x=2
73.
(x — 6) + log 2 (x — 4)— log, x = 2
(x — 6)(x — 4)
log, .
=2
(x —6)(x — 4) = 2,
71.
2 log, (x+ 4) = log, 9 + 2
2 log, (x + 4) = 2 +2
x 2 —10x + 24 = 4x
2 log, (x + 4) = 4
x 2 —14x + 24 = 0
log, (x + 4) = 2
(x — 12)(x — 2) = 0
3 2 =x+4
x —12 = 0
x— 2= 0
9=x+4
x = 12
x=2
5=x
The solution set is { 12} since loge (2-6) =
log, (-4) is not possible.
74. log, (x —3) + log, x — log, (x + 2) = 2
log,
(x — 3)x
(x+2)
=2
22
—x
2 — 3x
x+ 2
4(x +2) = —3x
4x + 8 = —3x
0 = — 7x — 8
0 = (x + I)(x — 8)
x+1 = 0
x —8 = 0
x = —1
x=8
log2 (-1-3) = log 2(-4) does not exist, so the solution set is { 8}.
76.
75. log(x + 4) = log x + log 4
log(5x + 1) = log(2x + 3) + log2
log(x + 4) = log 4x
log(5x + 1) = log(4x + 6)
x + 4 = 4x
5x+1=4x+6
4 = 3x
x=5
4
77.
3
This value is rejected. The solution set is
log(3x — 3) = log(x + 1) + log 4
log(3x — 3) = log(4x + 4)
4
3x-3=4x+4
—7 = x
This value is rejected. The solu on set
Is {
449
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earson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
N o portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Chapter 3
ISM: Precalcultis
78. log(2x -1) Iog(x + 3) + log3
log(2x -1) = log(3x +9)
2x-1=3x+9
-10 = x
This value is rejected. The solution set is { .
83. 2logx-log 7 = log112
log x2 - log 7 = log 112
22
log- = log112
7
x2
7=112
79. 2 log x = log 25
x2 = 784
x = ±28
-28 is rejected. The solution set is {28}.
log x2 = log 25
x2 = 25
x = ±5
-5 is rejected. The solution set is {5}.
84. log(x- 2) + log5 = log100
log(5x -10) = log100
5x-10=100
5x =110
x=22
80. 3 log x = log125
log x3 = log125
x' =125
x=5
85. log x +log(x +3) = log 10
log(x2 +3x) = log 10
x 2 ÷ 3x =10
x2+3x-I0
(x + 5)(x - 2) = 0
x = -5 or x = 2
-5 is rejected. The solution set is {2}.
81. log(x + 4)- log 2 = log(5x +1)
x +4
log
- log(5x +1)
x +4
— 5x +1
2
x+4=10x+2
-9x = -2
2
9
x 0.22
86. log(x + 3) + log(x - 2) = log14
log(x2 + x - 6) = log14
x2 + x - 6 =14
x2 +x- 20 =0
(x + 5)(x- 4) = 0
x = -5 or x = 4
-5 is rejected. The solution set is {4).
82. log(x + 7)- log3 = log(7x + 1)
x +7
log— log(7x + 1)
3
x +7
— 7 x +1
3
x +7 =21x +3
-20x = -4
1
x
87. in(x 4) + In(x +1) = ln(x -8)
In(x2 -3x -4) = ln(x -8)
x2 -3x-4 =x-8
x2 -4x+4=0
(x 2)(x - 2) = 0
x=2
2 is rejected. The solution set is { }.
x.0.2
450
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• trInnr Saddle River, NJ. All rights reserved. Thls material is
Eta
Chapter 3
JSM: Precaleului
92. 3 x+2 . 3x = 81
3 (,42)4: = 34
88. log,(x -1)- log, (x +3) = log, ( I)
log,
3 24+2 = 34
x-1
- log, (1)
x +3
x -1
1
x+3 x
x2 - x = x + 3
2x+2=4
2x = 2
x=I
93. 211nx1-6 =0
x2 - 2x -3 = 0
=0
x -1 or x = 3
-1 is rejected. The solution set is {3}.
(x +1)(x -3)
21In xl = 6
1111
3
x =3
89. In(x - 2) - In(x +3) =111(x -1)- Iti(x + 7)
x -1
In x-2
= In
x+7
xt3
x -2 x -1
x +3 x + 7
(x - 2)(x + 7) = (x +3)(x -1)
x
x
or In x = -3
x = e-3
x 0.05
e3
20.09
94. 3Ilog xl- 6 = 0
3Ilogx1= 6
llog xi = 2
x2 + 5x -14 = x2 + 2x -3
log x = 2
or
x=102
x=100
95.
90. In(x - 5) - In(x +4) = In(x -1) -113(x + 2)
log x = -2
x=10-2
x=0.01
3' 2 = 45
In 3? = In 45
x - 5 _ x -1
x+2
x+4
x -5 _ x -1
x+4 x+2
(x -5)(x + 2) = (x+ 4)(x -1)
x2 In 3= In 45
x2 7.4
1n45
In 3
x =±1111145
Ina
x 2 -3x -10 = x 2 +3x - 4
-6x =
x = -1
-1 is rejected. The solution set is }.
96.
±1.86
5' 2 = 50
In 5.e = In 50
x2 In 5= In 50
2 MO
x
ln 5
91. 5 2x . 54x =125
52x+4x = 53
5 6K = 53
50
x = ±1 111 ± 1 . 5 6
In 5
6x=3
1
x=2
451
2007 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.
ISM: Precalculus
Chapter 3
97.
In(2x+1)+In(x —3)— 21n x = 0
x=
In(2x +1) + In(x —3)— In x2 = 0
(2x +1)(x-3) —0
In
x2
(2x +1)(x —3) — co
1
x
2X2 - 5x —3
2
x=
x=
—1
x
x
2X 2 - 5x —3 = x2
—b
x
x—
x=
2a
—(5) ±
(5) 2 —4(1)(-3)
2(1)
—Stift
2
2
g30.54
5— ff7
2
a5.54 (rejected)
—4ac
The solution set is
2a
x=
x
tie
x=
x2 —5x-3 = 0
—4ac 2
x +5x —3= 0
—5+
I-5+
}
2
—(-5)±4(-5)2 —4(1)(-3)
2(1)
99.
5)0-12 = (52 )2x
5±VT7
2
5+ VT7
2
= 5—
5.54
x 2 — 12 = 4x
—4x—l2 = 0
0.54 (rejected)
2
The solution set is
{x-6}(x+2)=0
Apply the zero product property:
x-6=0 or x+2=0
x = —2
x= 6
The solutions are —2 and 6, and the solution set is
{-2,
.
{5 + /F7}
2
98. In 3 —1n(x +5)— In x = 0
In
3— 0
x(x +5)
et)
=
1=
100.
3
x(x+ 5)
3
3x2-12 = 92x
3x 2 -12 = (32 )2x
3x 2 -12
x(x +5)
x(x+5)=3
2
x -F 5x = 3
34x
x2-12=4x
x2 —4x --12 = 0
{x-6){x+2)=0
Apply the zero product property:
x-6=0 or x+2=0
x = —2
x=6
The solutions are —2 and 6, and the solution set is
{-2,
.
x2+5x-3=0
452
2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl y exist'
No portion of this material may be reproduced, In any form or by any means, without permission In writing from the publisher.
Chapter 3
ISM: Precalculus
104. a. In 2000 t = 0
101. 25 = 6e 12 nx
A(0) = 15.9e
625
__
6
A(0) =15.9e0
A(0) =15.9
In 2000, the population was 15.9 million.
In el2.77x
In 25 =12.77x
6
25
6 _x
12.77
0.112 x
A blood alcohol level of about 0.11 corresponds
to a 25% risk of a car accident.
102.
0.0235(0)
25 _ 6 12 77x
b.
50 = 6e12.77x
e 12.77x =
25
in 812.77x .111(251
C3 j
12.77x =
3
in ( 25)
3
x=
x0.17
12.77
A blood alcohol level of about 0.17
corresponds to a 50% risk of a car
accident.
103. a.
b.
105.
19.2 =15.9e 0.0235t
19 .2 = 0.0235/
15.9
In 19.2 _• In e 0 0235i
15.9
19.2
= 0.0235:
In
15.9
In19.2
15.9 _ t
0.0235
8xt
The population will reach 19.2 million
about 8 years after 2000, in 2008.
0 0575)4'
20,000 =12,500(1+ '
4
12,500(1.014375)4' = 20,000
(1.014375)4' =1.6
ln(I.014375) 4' = 1n1.6
4t ln(1.014375) = 1n1.6
In1.6
t=
41n1.014375
A = 18.9e° ®5(c)
A =18.9 million
8.2
8.2 years
19.6 =18.9e44°551
106.
19.6
oma'
=c
18.9
= e o oossr
18.9
19.6
In
= 0.0055t
18.9
19.6
In
18.9 _ t
0.0055
6.6 x t
In 2007 the population of New York will
reach 19.6 million.
15,000 = 7250(1+
0.1026112'
7250(1.005416667) 121 =15,000
(1.005416667) 121 =
29
i6O)
In(1.005416667) 121 = ht C29
)
12t Ina 00541667) =
r 60)
C29
60)
C29)
x 11.2 years
t=
12In1.00541667
453
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N o portion of this material may be reproduced, In any form or by any means, without permission In writing from the publisher.
ISM: Precalculus
Chapter 3
107.
1400 =1000(1+
r 3002
360
110. 12,000 = 8000e' 2
e =1.5
In e2r = In 1.5
2r =1n1.5
M1. a<
r5 0.203
2
20.3%
720
=1.4
(1+360)
720
In(1+
=1111.4
360 )
7201n(1+-2---1=1111.4
360
r ) 1n1.4
1+360
111(
720
ln(1+14.0)
= e (b11.4)/ 720
e
1+ r = tiAnno _ 1
360
111. accumulated amount = 3(2350)- 7050
7050 = 2350e1
lir = 3
In e7' = 1113
7r =1113
r = In 3 = 0.157
r = 360(e ""2°) -1
15.7%
Rd 0.168
16.8%
112. 25,000 =17, 425eum23'
e 00423. = 1000
697
In emu ra. (1000')
ll
697 )
0.04251 =111(1000)
697 )
hir1000)
697 ) = 8.5 years
t=
0.0425
TWO)
10t 9000= 5000(1+
MO
1440
1.8
(1
+360
111+
1440
=101.8
- L360
1440111( 1+-21-=1111.8
360
ht 1+
r )=In1.8
113. a.
360) 1440
e in0+S“) = e0n1.8)/1440
The function result of 64.3% models the
actual value of 63% very well.
1+ r = "1.8)/1440 -1
360
r
f (x) = 13.4 + 46.31n x
f (3)= 13.4 + 46.31n3 = 64.3
b.
eQ41.8)/1440 -1
360
r =360(0")/14'w -1 = 0.147
f (x) =13.4 + 46.31n x
96 =13.4+ 46.31n x
82.6 = 46.31n x
82.6
- x
46.3
14.7%
109. accumulated amount = 2(8000) = 16,000
16,000 = 8000e° m'
e°'"' 2
e0.081 = 2
82.6
x=
x =6
The function predicts that 96% of email
will be spam 6 years after 2000, or 20g
0.08t =1n 2
hit
t=
0.08
t = 8.7
The amount would double in 8.7 years.
454
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material Is protected under all copyright laWe
__
-.
mnroduced. In an
means. without Permission In writing from the
pd
123.
4 —2.6in x
24.3
um) result of 24.3% models the
alue of 24% very well.
34-2.0/141.
=34 –2.61n x
– 2.6In x
The intersection point is (2, 8).
Verify: x = 2
=. –2.61n x
2"' =8
=In x
2 2'' = 2
6
23 =8
8=8
The solution set is {2 } .
em
x Pa 69
lie function predicts that children under
'8 will decline to 23% of the total
opulation 69 years after 1969, or 2038.
124.
$2x=-45
10
10
ix".5
x=2" = 2.8
Jy. half the students recall the important
tpres of the lecture after 2.8 days.
50)
I
to
–10
{1)
The intersection point is (1, 9).
Verify x = 1:
3" = 9
3 iii = 9
95 –30 log2 x
Olog2 x = 95
95
3 2 =9
9=9
The solution set is 11.
log, x = —
30
x
9.0 = x
(9.0, 0)
125.
7. 2.4 = – log x
log x = –2.4
x= 10 .24 g--; 0.004
The hydrogen ion concentration was 10--2.4,
approximately 0.004 moles per liter.
The intersection point is (4, 2).
Verify: x = 4
log,(4 4 – 7) = 2
log, 9 = 2
2=2
The solution set is {4}.
118. 4.2 = – log x
log x = –4.2
10-42 0.00006
The hydrogen ion concentration was 10-42,
approximately 0.00006 moles per liter.
x
455
0,2007 P earson 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 3
126.
ISM: Precut(
129.
10
10
10
—10
The intersection point is (31 , 2).
There are 2 points of intersection, approxima
(-1.391606, 0.21678798) and
(1.6855579, 6.3711158).
Verify x z -1.391606
2x+3
1.991606 rz
3A 1.391606) + 3
0.2167879803 z- 0.216788
Verify x z 1.6855579
3 3` = 2x + 3
3 1 6855579 "^t 2(1.6855579) + 3
6.37111582 z 6.371158
The solution set is (-1.391606, 1.6855579).
1
Verify: x = —
3
log, (3 .11 -2) = 2
3
log, (11-2) = 2
log, 9 = 2
2=2
The solution set is M.
127.
130.
0
There are 2 points of intersection,
approximately
(-1.291641, 0.12507831) and
(1.2793139,7.8379416).
Verify:x z -1.291641
=3x+4
5 -1291641 = 3(-1.291641)+4
0.1250782178 z 0.125077
Verify:x =1.2793139
25
19
2
-2
The intersection point is (2, 1).
Verify: x = 2
log(2 + 3) + log 2 =1
log5+ log2 =1
log(5 . 2) = 1
log10 =1
1=1
The solution set is {2}.
128.
10
5' 2793139 = 3(1.2793139)+4
7.837941942 z 7.8379417
The solution set is (-1.291641, 1,2793139).
21
15
131
The intersection point is (20, 2).
Verify x = 20:
log (x - 15) + log x = 2
log (20 - 15) + log 20 = 2
log 5 + log 20 = 2
log 100 = 2
100 = 102
100 = 100
The solution set is (20).
0
500
27
As the distance from the eye increases,
a
barometric air pressure increases, leveling
about 30 inches of mercury.
456
rrn ,nn7 Paarcnn
Fduration. Inc,. Uo per Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as
•
29 = 0 48 In(x +1) +27
Chapter 3
1
True; x = — In 3?
c.
÷=2
x = in y
0.24
110 +I)
d
e h.
elny
ono ern.
y
eX1,
x +1
d.
63.5
x e rg" — 1
False; The equation x 1 ° = 5,71 has no
variable in an exponent so is not an
exponential equation.
barometri c air pressure is 29 inches of
cury at a distance of about 63.5 miles from
eye of a hurricane.
(c) is true
136. Account paying 3% interest:
A = 4000 I+
•
0.03
I
/
Account paying 5% interest:
05
A = 2000I+ 0.
1
■.
/
The two accounts will have the same balance
when
40000.03)' = 20000.05)'
he point of intersection is approximately
3.5, 29),
a
(1.03)' = 0.5(1.05)'
(1.03"
0.5
,1.05,
0
15
0
1111) -1132
)
= In 0.5
1.05)
When P = 70,
minutes.
Verify;
1= 7.9,
so it win take about 7.9
t ln
03‘
J.05,
= In 0.5
t—
70 = 45e-9992)99)
In 0.5
36
(1 3\
In 11
v1051
The accounts will have the same balance in
about 36 years.
70 .‘). 70.10076749
The runner's pulse will be 70 beats per minute
after about 7.9 minutes.
4. 3000
137.
(In x) 2 = In .1-9
(In x) 2 = 21.n x
(lnxf —21nx=0
100
In x(ln x —2) = 0
Inx=2
e lnx e 2 or
An adult female elephant weighing 1800
kilograms is about 20 years old.
35. a,
b.
=0
x
The solution set is 11, e2}.
False; log(x + 3) = 2 means x + 3 = 109
False; log(7x +3)— log(2x +5) = 4 means
log
7x +3
2x +5
= 4 which means
7x+3
or
2x+ 5
457
Q07 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright
laws as they
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.currently exist.
ISM: Preealculas
Chapter 3
Check with graphing utility:
139. In(In x) 0
el as) =
enx = el
x=e
The solution set is {e}.
There are two points of intersection: (1, 0) and
approximately (7.3890561, 4). Since
7.3890566099, the graph verifies x = 1 and
x = e2, so the solution set is {1, e2} as
determined algebraically.
The graph of In(In(x)) crosses the graph y =
approximately 2.718.
138. (log x)(2 log x + 1) = 6
2(logx)2 + log x — 6 = 0
(2 logx — 3Xlogx + 2) = 0
2logx-3= 0 or logx+2= 0
log x = —2
2 log x = 3
3
log x = —
2
x =10-2
Section 3.5
Check Point Exercises
1. a. Use the exponential growth model
A = Aoe m with 1990 corresponding
when the population was 643 millio
A = 643e0
Substitute t = 2000 — 1990 = 10 w1t0
population was 813 million, so A
find k.
813 = 643e°
x=101'
x = 10.1173
The solution set is{104161.
100'
Check by direct substitution:
Check:x =10\5 =103/2
(log x)(2log x +1) = 6
(log10312 )(21og10312 + 1) = 6
813 = e k10
643
83
= ine°
43
813
In— =10k
643
813
al-643 — k
10
0.023 as k
So the exponential growth
A = 643e°023
(-1)(2 .1
2 +1)= 6
(1) (3 +1) = 6
• (2
2) (4) = 6
6=6
458
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No Donlon of this material may be reproduced, in any form or by any means, without permission In writiiltl