C H A P T E R 3 Exponential and Logarithmic Functions

CHAPTER 3
Exponential and Logarithmic Functions
Section 3.1 Exponential Functions and Their Graphs . . . . . . . . . 265
Section 3.2 Logarithmic Functions and Their Graphs . . . . . . . . 273
Section 3.3 Properties of Logarithms . . . . . . . . . . . . . . . . . 281
Section 3.4 Exponential and Logarithmic Equations . . . . . . . . . 289
Section 3.5 Exponential and Logarithmic Models . . . . . . . . . . 303
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333

CHAPTER 3
Exponential and Logarithmic Functions
Section 3.1 Exponential Functions and Their Graphs
■ You should know that a function of the form where is called an exponential
function with base a.
■ You should be able to graph exponential functions.
■ You should know formulas for compound interest.
(a) For n compoundings per year:
(b) For continuous compoundings: A Pert r
A P1 n
.
nt
f x a a > 0, a 1,
.
x ,
Vocabulary Check
1. algebraic 2. transcendental 3. natural exponential; natural
n nt
4. 5. A Pert r
A P1
1. f5.6 3.4 5.6 946.852 2. fx 2.3 x 2.3 32 3.488 3. f 5 0.006
4. f x 2 3 5x 2 3 50.3 0.544 5.
gx 50002 x 50002 1.5 6.
1767.767
f x 2001.2 12x
2001.2 1224
1.274 10 25
7. fx 2
Increasing
Asymptote: y 0
Intercept: 0, 1
Matches graph (d).
x 8. f x 2 rises to the right.
Asymptote: y 1
Intercept: 0, 2
Matches graph (c).
x 1 9. fx 2
Decreasing
Asymptote: y 0
Intercept: 0, 1
Matches graph (a).
x
10. fx 2 rises to the right.
Asymptote: y 0
x2 11.
Intercept:
0, 1
4
Matches graph (b).
f x 1 2 x
x 2
1 0 1 2
f x
4 2 1 0.5 0.25
Asymptote: y 0
5
4
3
2
1
y
−3 −2 −1
−1
1
2
3
x
265

266 Chapter 3 Exponential and Logarithmic Functions
12.
14.
16.
17.
19.
f x 1 2 x
2x x 2 1 0 1 2
f x
0.25
Asymptote:
5
4
3
2
y
−3 −2 −1
−1
f x 6 x
Asymptote:
5
4
3
2
1
y
−3 −2 −1
−1
y 0
1
1
0.5
y 0
2
2
3
3
1 2 4
x 2 1 0 1 2
f x
f x 4 x3 3
0.028 0.167 1 6 36
x 1 0 1 2 3
f x
3.004
Asymptote: y 3
3.016
x
x
3.063
3.25
7
6
5
4
2
1
13.
15.
f x 6 x
x 2 1 0 1 2
f x
−3 −2 −1 1 2 3 4 5
gx 3x4 f x 3x , 18.
Because gx f x 4, the graph of g can be obtained
by shifting the graph of f four units to the right.
gx 5 2x f x 2x , 20.
Because gx 5 f x, the graph of g can be obtained
by shifting the graph of f five units upward.
4
y
Asymptote:
5
4
3
1
y
−3 −2 −1
−1
f x 2 x1
Asymptote:
−3 −2 −1
−1
36 6 1 0.167 0.028
5
4
3
2
1
y
y 0
1
y 0
1
2
x 2 1 0 1 2
f x
x
0.125
2
3
0.25
f x 4 x , gx 4 x 1
3
x
0.5
x
1 2
Because gx f x 1, the graph of g can be obtained
by shifting the graph of f one unit upward.
f x 10
Because gx f x 3, the graph of g can be
obtained by reflecting the graph of f in the y-axis and
shifting f three units to the right. (Note: This is equivalent
to shifting f three units to the left and then reflecting the
graph in the y-axis.)
x , gx 10x3

f x 7 2 x
, gx 7 2 x6
21.
Because gx fx 6, the graph of g can be
obtained by reflecting the graph of f in the x-axis and
y-axis and shifting f six units to the right. (Note: This is
equivalent to shifting f six units to the left and then
reflecting the graph in the x-axis and y-axis.)
23.
y 2 x2
−3
3
−1
3
24.
y 3 x 25.
3
−3 3
−1
Section 3.1 Exponential Functions and Their Graphs 267
22.
gx 0.3
gx fx 5, hence the graph of g can be obtained
by reflecting the graph of f in the x-axis and shifting the
resulting graph five units upward.
x f x 0.3 5
x ,
fx 3 x2 1 26.
27. f3 4 e34 0.472 28. fx ex e3.2 24.533 29. f 10 2e510 3.857 1022 30.
33.
35.
f x 1.5e 12x 31. f 6 5000e 0.066 7166.647 32.
1.5e 120 1.956 10 52
f x e x
x 2 1 0 1 2
f x
Asymptote:
0.135
5
4
3
2
1
y
−3 −2 −1
−1
f x 3e x4
x
f x
8
0.055
Asymptote:
y 0
1
y 0
0.368
2
7
−8 −7 −6 −5 −4 −3 −2 −1 1
3
0.149
8
7
6
5
4
3
2
1
y
x
x
1 2.718 7.389
6
0.406
5
1.104
4
3
34.
36.
−1
4
0
f x e x
Asymptote:
5
4
3
2
1
y
−3 −2 −1
−1
y 0
1
2
5
3
f x 250e 0.05x
x
y 4 x1 2
−6 3
3
−3
250e 0.0520 679.570
x 2 1 0 1 2
f x
f x 2e 0.5x
Asymptote:
6
5
4
3
2
1
−3 −2 −1
−1
7.389 2.718 1 0.368 0.135
x 2 1 0 1 2
f x
5.437 3.297 2 1.213 0.736
y
y 0
1
2
3
4
x

268 Chapter 3 Exponential and Logarithmic Functions
37.
39.
42.
45.
48.
51.
f x 2e x2 4
x 2 1 0 1 2
f x
Asymptote:
−3 −2 −1
9
8
7
6
5
3
2
1
y
4.037
1
2
y 4
3
4
4.100
5
6
7
x
4.271
4.736
y 1.08 5x 40.
−7
x 1 3
1 1 1 3 x1 27 46.
3 x1 3 3
5 x1
5 x1
5 x1
x 2
5 3
x 1 3
x 4
7
−1
1 5 3
5
st 3e 0.2t 43.
20
−16 17
−2
125 49.
6
2 x3 2 4
x 3 4
x 7
38.
−4 8
f x 2 e x5
Asymptote:
−1
x 0 2 4 5 6
f x
8
7
6
5
4
3
1
y
1
2
2.007 2.050 2.368 3 4.718
3
y 2
2 x3 16 47.
e 3x2 e 3 50.
3x 2 3
3x 1
x 1
3
e x2 3 e 2x 52.
x 2 3 2x
x 2 2x 3 0
x 3x 1 0
x 3 or x 1
y 1.08 5x 41.
6
−2
gx 1 e x 44.
−3
4
0
3
4
5
6
7
e x2 6 e 5x
x 2 6 5x
x 2 5x 6 0
x 3x 2 0
x 3 or x 2
8
−10
x
st 2e 0.12t
22
0
hx e x2
4
−2 4
0
2 x2 1
32
2 x2 2 5
x 2 5
x 3
e 2x1 e 4
2x 1 4
2x 5
x 5
2
23

53.
54.
55.
P $2500, r 2.5%, t 10 years
Compounded n times per year:
Compounded continuously: A Pe rt 2500e 0.02510
58. A Pe rt 12,000e 0.06t
r
A P1 n nt
0.025
25001
n 10n
n 1 2 4 12 365
Section 3.1 Exponential Functions and Their Graphs 269
Continuous
Compounding
A $3200.21 $3205.09 $3207.57 $3209.23 $3210.04 $3210.06
P $1000, r 4%, t 10 years
Compounded n times per year:
0.04
A 10001
n 10n
Compounded continuously: A 1000e 0.0410
n 1 2 4 12 365
Continuous
Compounding
A $1480.24 $1485.95 $1488.86 $1490.83 $1491.79 $1491.82
P $2500, r 3%, t 20 years
Compounded n times per year:
r
A P1 n nt
0.03
25001
n 20n
Compounded continuously: A Pe rt 2500e 0.0320
n 1 2 4 12 365
Continuous
Compounding
A $4515.28 $4535.05 $4545.11 $4551.89 $4555.18 $4555.30
56. P $1000, r 6%, t 40 years
Compounded n times per year:
0.06
A 10001
n 40n
Compounded continuously: A 1000e 0.0640
n 1 2 4 12 365
Continuous
Compounding
A $10,285.72 $10,640.89 $10,828.46 $10,957.45 $11,021.00 $11,023.18
57. A Pe rt 12,000e 0.04t
t 10 20 30 40 50
A $17,901.90 $26,706.49 $39,841.40 $59,436.39 $88,668.67
t 10 20 30 40 50
A $21,865.43 $39,841.40 $72,595.77 $132,278.12 $241,026.44

270 Chapter 3 Exponential and Logarithmic Functions
59. A Pe rt 12,000e 0.065t
t 10 20 30 40 50
A $22,986.49 $44,031.56 $84,344.25 $161,564.86 $309,484.08
60. A Pe rt 12,000e 0.035t
61.
64.
t 10 20 30 40 50
A $17,028.81 $24,165.03 $34,291.81 $48,662.40 $69,055.23
A 25,000e 0.087525 62.
$222,822.57
4
p 50001
4 e
(a) 1200
0.002x
0
0
(b) When x 500:
p 50001
2000
4
4 e0.002500 $421.12
(c) Since 600, 350.13 is on the graph in part (a), it
appears that the greatest price that will still yield a
demand of at least 600 units is about $350.
In 2000, and the population is given by
P10 152.26e0.003910 t 10
146.44 million.
(c) In 2010, and the population is given by
P20 152.26e0.003920 t 20
140.84 million.
A 5000e 0.07550 63. C10 23.951.04 10 $35.45
$212,605.41
65.
Vt 100e 4.6052t
(a)
(b)
V1 10,000.298 computers
V1.5 10,004.472 computers
(c) V2 1,000,059.63 computers
66. (a)
Since the growth rate is negative,
the population is decreasing.
(b) In 1998, and the population is given by
P8 152.26e0.00398 P 152.26e
0.0039 0.39%,
t 8
147.58 million.
0.0039t 67. Q 25
(a) Q0 25 grams
(b) Q1000 16.21 grams
(c) 30
1
2 t1599
68.
(a) When t 0: Q 101 Q 101 2 t5715
(b)
2 05715
101 10 grams
When t 2000: Q 101 2 20005715
7.85 grams
(c)
Mass of 14 C (in grams)
12
10
8
6
4
2
Q
4000 8000
Time (in years)
0 5000
0
t

69.
y
(a)
70. (a)
100
1 7e 0.069x
110
0 120
0
Section 3.1 Exponential Functions and Their Graphs 271
71. True. The line is a horizontal asymptote for the
graph of fx 10x y 2 72. False, e e is an irrational number.
2.
271,801
99,990 .
73.
Atmospheric pressure
(in pascals)
fx 3 x2 74.
3 x 3 2
3 x 1
3 2
1
9 3x
hx
Thus, fx gx, but fx hx.
75. and f x 164x fx 164
x 76.
4 2x
1 4 2x
1
120,000
100,000
4 2 4 x
4 x2
gx
80,000
60,000
40,000
20,000
Thus, fx gx hx.
P
5 10 15 20
Altitude (in km)
25
h
162 2 x
162 2x
hx
(b)
x Sample Data Model
0 12 12.5
25 44 44.5
50 81 81.82
75 96 96.19
100 99 99.3
(b) p 107,428e0.150h 107,428e 0.1508
32,357 pascals
gx 2 2x6
2 2x 2 6
642 2x
642 2 x
644 x
hx
Thus, gx hx but gx fx.
fx 5 x 3
gx 5 3x 5 3 5 x
hx 5 x3 5 x 5 3
Thus, none are equal.
(c) When x 36:
y
100
63.14%.
1 7e0.06936 2
100
(d) 100
when
3 1 7e
x 38 masses.
0.069x

272 Chapter 3 Exponential and Logarithmic Functions
77.
78. (a)
81.
y 3 x and y 4 x
−2
3
y = 4x −1
fx x 2 e x (b)
5
−2 7
−1
Decreasing:
Increasing:
Relative maximum: 2, 4e
Relative minimum: 0, 0
2 , 0, 2,
0, 2
y 2 25 x 2
y ±25 x 2
gx x2 3x
x 2 y 2 25 82.
−2
6
−2
Decreasing: 1.44,
Increasing: , 1.44
Relative maximum: 1.44, 4.25
0.5
79. fx 1 and gx e (Horizontal line)
0.5 80. The functions (c) and (d) 2 are exponential.
x
3x f
g
2
1
−1
4
x x
−3 3
0
y
As x → , fx → gx.
As x → , fx → gx.
1
y = 3 x
83. f x
Vertical asymptote: x 9
2
9 x
Horizontal asymptote: y 0
x
f x
11
1
10
2
2
x
8
(a)
x 2 1 0 1 2
3 x
4 x
1
9
1
16
1
3
1
4
4 x < 3 x when x < 0.
(b) 4 x > 3 x when x > 0.
7
2 1
1 3 9
1 4 16
−18 −15 −6
12
9
6
3
−3
−3
−6
−9
y
3
x
10
x 2
y x 2 and y x 2, x ≥ 2
y
x y 2

84. f x 7 x
Domain: , 7
x
y
9
2
Section 3.2 Logarithmic Functions and Their Graphs
Section 3.2 Logarithmic Functions and Their Graphs 273
■ You should know that a function of the form y loga x, where a > 0, a 1, and x > 0, is called
a logarithm of x to base a.
■ You should be able to convert from logarithmic form to exponential form and vice versa.
y log a x ⇔ a y x
■ You should know the following properties of logarithms.
(a)
(b)
(c)
(d) a Inverse Property
(e) If loga x loga y, then x y.
■ You should know the definition of the natural logarithmic function.
log loga a
a x x
x x since ax ax loga a 1 since a
.
1 loga 1 0 since a
a.
0 1.
log e x ln x, x > 0
■ You should know the properties of the natural logarithmic function.
(a)
(b)
(c)
(d) e Inverse Property
(e) If ln x ln y, then x y.
■ You should be able to graph logarithmic functions.
ln x ln e
x
x x since ex ex ln e 1 since e
.
1 ln 1 0 since e
e.
0 1.
Vocabulary Check
3 6 7
4 3 2 1 0
1. logarithmic 2. 10 3. natural; e
4. a 5. x y
loga x x
1. log 4 64 3 ⇒ 4 3 64 2. log 3 81 4 ⇒ 3 4 81 3. log 7 1
49 2 ⇒ 72 1
49
4. log 1
1000 3 ⇒ 103 1
1000
5. log 32 4 2
5 ⇒ 3225 4 6. log 16 8 3
4 ⇒ 1634 8
7. log 36 6 1
2 ⇒ 3612 6 8. log 8 4 2
3 ⇒ 823 4 9. 5 3 125 ⇒ log 5 125 3
10. 8 2 64 ⇒ log 8 64 2 11. 81 14 3 ⇒ log 81 3 1
4
6
4
2
−4
−6
y
−4 −2
−2
2
4 6 8
x
85. Answers will vary.
12. 9 32 27 ⇒ log 9 27 3
2

274 Chapter 3 Exponential and Logarithmic Functions
13. 6 2 1
36 ⇒ log 6 1
36 2 14. 4 3 1
64 ⇒ log 4 1
64 3 15. 70 1 ⇒ log 7 1 0
16. 103 0.001 ⇒ log10 0.001 3 17.
f16 log2 16 4 since 24 fx log2 x
16
22.
25.
28.
31.
gx log b x 23.
gb 3 log b b 3 3
b 3 b 3
f 12.5 1.097
since
f x log x 26.
Since 1.5 log1.5 1 0.
0 1,
f4 5 log4 f x log x 24.
5 0.097
f 1
500 log 1
f x log x
f 75.25 1.877
500
2.699
f x log x 27. since 34 34 log3 34 4
log1.5 1 29. since 1 log 1 .
30.
fx log 4 x
Domain: x > 0 ⇒ The domain is 0, .
x-intercept:
1, 0
Vertical asymptote:
x 0
y log 4 x ⇒ 4 y x
x 1 4 2
f x
1
4
1
33. y log3 x 2
Domain: 0,
x-intercept:
log 3 x 2 0
0 1
2 log 3 x
3 2 x
9 x
The x-intercept is 9, 0.
Vertical asymptote:
y log 3 x 2
1
2
log 3 x 2 y ⇒ 3 2y x
x 27 9 3 1
x 0
1
3
y 1 0 1 2 3
2
1
−1 1 2 3
6
4
2
−2
−4
−6
y
−1
−2
y
2 4 6 8 10 12
x
x
32. gx log6 x
Domain: 0,
x-intercept:
1, 0
Vertical asymptote:
y log 6 x ⇒ 6 y x
1
6
9 log 9 15
Since 9log aloga 9 15 15.
x x,
x 0
x 1 6 6
1
2
y 1 0 1
34. hx log4x 3
Domain:
The domain is 3, .
x-intercept:
log 4x 3 0
4 0 x 3
1 x 3
4 x
The x-intercept is 4, 0.
Vertical asymptote:
−2
−4
y log 4x 3 ⇒ 4 y 3 x
1
4
18.
x 3 > 0 ⇒ x > 3
x 3 4 7 19
y 1 0 1 2
fx log 16 x
f4 log 16 4 1
2
19.
f1 log7 1 0 since 70 fx log7 x 20.
1
since 101 fx log x 21. gx loga x
f10 log 10 1 10
ga 2 log a a 2
6
4
2
2
1
−1 1 2 3
−1
−2
x 3 0 ⇒ x 3
y
y
since 16 12 4
2 by the Inverse Property
2 4 6 8 10
x
x

35. fx log6x 2
Domain: x 2 > 0 ⇒ x > 2
The domain is 2, .
x-intercept:
0 log 6x 2
0 log 6x 2
6 0 x 2
1 x 2
1 x
The x-intercept is 1, 0.
Vertical asymptote: x 2 0 ⇒ x 2
y log 6x 2
y log 6x 2
6 y 2 x
x 4
f x
1
1
1 5
6
0 1 2
1 35
36
4
2
−2
−4
y
6
Section 3.2 Logarithmic Functions and Their Graphs 275
x
36.
y log 5x 1 4
Domain:
The domain is 1, .
x-intercept:
log 5x 1 4 0
log 5x 1 4
626
625 x
The x-intercept is
Vertical asymptote:
x 1 > 0 ⇒ x > 1
5 4 x 1
1
625 x 1
626 625 , 0.
x 1 0 ⇒ x 1
y log 5x 1 4 ⇒ 5 y4 1 x
6
5
4
3
2
1
y
2 3 4 5 6
x 1.00032 1.0016 1.008 1.04 1.2
y 1 0 1 2 3
37.
Domain:
The domain is
x-intercept:
The x-intercept is 5, 0.
Vertical asymptote:
x
0 ⇒ x 0
5
The vertical asymptote is the y-axis.
x
1 ⇒ x 5
5 x
log
100
5 x
y log
x
> 0 ⇒ x > 0
5
0, .
y
4
0 5
2
x
−2
4 6 8
−4
x
5 x 1 2 3 4 5 6 7
y 0.70 0.40 0.22 0.10 0 0.08 0.15
38.
y logx
Domain: x > 0 ⇒ x < 0
The domain is , 0.
x-intercept:
logx 0
The x-intercept is 1, 0.
Vertical asymptote:
10 0 x
1 x
x 0
y logx ⇒ 10 y x
x
1
100
1
10
y 2 1 0 1
−3 −2 −1 1
2
1
−1
−2
y
1
x
10
x

276 Chapter 3 Exponential and Logarithmic Functions
39. fx log3 x 2 40. fx log3 x
Asymptote: x 0
Asymptote: x 0
Point on graph: 1, 2
Point on graph: 1, 0
Matches graph (c).
Matches graph (f).
The graph of f x is obtained by shifting the graph of gx
upward two units.
f x reflects gx in the x- axis.
41.
43.
fx log 3x 2 42.
Asymptote:
Point on graph:
x 2
1, 0
Matches graph (d).
The graph of f x is obtained by reflecting the graph of
gx about the x- axis and shifting the graph two units to
the left.
fx log 31 x log 3x 1 44.
Asymptote:
Point on graph:
x 1
0, 0
Matches graph (b).
The graph of f x is obtained by reflecting the graph of
gx about the y- axis and shifting the graph one unit to
the right.
45. ln 1
2 0.693 . . . ⇒ e0.693 . . . 1
2
fx log 3x 1
Asymptote:
Point on graph:
x 1
2, 0
Matches graph (e).
f x shifts gx one unit to the right.
fx log 3x
Asymptote:
Point on graph:
x 0
1, 0
Matches graph (a).
f x reflects gx in the x axis then reflects that graph in
the y- axis.
46. ln 2
5 0.916 . . . ⇒ e0.916 . . . 2
5
47. ln 4 1.386 . . . ⇒ e 1.386 . . . 4 48. ln 10 2.302 . . . ⇒ e 2.302 . . . 10
49. ln 250 5.521 . . . ⇒ e 5.521 . . . 250 50. ln 679 6.520 . . . ⇒ e 6.520 . . . 679
51. ln 1 0 ⇒ e 0 1 52. ln e 1 ⇒ e 1 e
53. e 3 20.0855 . . . ⇒ ln 20.0855 . . . 3 54. e 2 7.3890 . . . ⇒ ln 7.3890 . . . 2
55. e 12 1.6487 . . . ⇒ ln 1.6487 . . . 1
2
56. e 13 1.3956 . . . ⇒ ln 1.3956 . . . 1
3
57. e 0.5 0.6065 . . . ⇒ ln 0.6065 . . . 0.5 58. e 4.1 0.0165 . . . ⇒ ln 0.0165 . . . 4.1
59. e x 4 ⇒ ln 4 x 60. e 2x 3 ⇒ ln 3 2x
61. fx ln x 62. fx 3 ln x
f18.42 ln 18.42 2.913
f0.32 3 ln 0.32 3.418
63. gx 2 ln x
g0.75 2 ln 0.75 0.575
64.
g1 2 ln 1
gx ln x
2 0.693

65.
gx ln x 66.
ge by the Inverse Property
3 ln e3 3
Section 3.2 Logarithmic Functions and Their Graphs 277
gx ln x
ge 2 ln e 2 2
67.
ge by the Inverse Property
23 ln e23 2
gx ln x 68.
3
ge52 ln e52 5
gx ln x
2
69.
f x lnx 1
Domain:
The domain is 1, .
x- intercept:
0 lnx 1
e 0 x 1
2 x
x 1 > 0 ⇒ x > 1
The x-intercept
is 2, 0.
−1
−1
Vertical asymptote: x 1 0 ⇒ x 1
x
f x
1.5 2 3 4
0.69
71. gx lnx
73.
Domain:
The domain is , 0.
x-intercept:
e 0 x
1 x
The x-intercept is 1, 0.
3
2
1
−2
−3
0 0.69 1.10
Vertical asymptote: x 0 ⇒ x 0
x
0 lnx
gx
x > 0 ⇒ x < 0
0.5
0.69
1
2
y
1
2 3 4 5
−3 −2 −1 1
3
0 0.69 1.10
y 1 logx 1 74.
−1
2
−2
5
2
1
−2
y
x
x
70.
72.
hx lnx 1
Domain:
The domain is 1, .
x-intercept:
lnx 1 0
e 0 x 1
1 x 1
0 x
The x-intercept is 0, 0.
Vertical asymptote:
x 1 > 0 ⇒ x > 1
y lnx 1 ⇒ e y 1 x
−2 2 4 6 8
x 1 0 ⇒ x 1
x 0 1.72 6.39 19.09
y 0 1 2 3
1
0.39
2
fx ln3 x
Domain: 3 x > 0 ⇒ x < 3
The domain is , 3.
x-intercept:
ln3 x 0
e 0 3 x
1 3 x
2 x
The x-intercept is 2, 0.
Vertical asymptote: 3 x 0 ⇒ x 3
y ln3 x ⇒ 3 e y x
x 2.95 2.86 2.63 2 0.28
y 3 2 1 0 1
fx logx 1 75. y1 lnx 1
2
−1 5
−2
0
3
−3
y
3
2
6
4
2
−2 −1 1 2 4
−1
−2
−3
y
9
x
x

278 Chapter 3 Exponential and Logarithmic Functions
76.
79.
81.
83.
85.
fx lnx 2 77.
3
−4 5
−3
log 2x 1 log 2 4 80.
x 1 4
x 3
log2x 1 log 15 82.
2x 1 15
x 7
lnx 2 ln 6 84.
x 2 6
x 4
lnx 2 2 ln 23 86.
x 2 2 23
x 2 25
x ±5
When x $1254.68:
1254.68
t 12.542 ln
20 years
1254.68 1000
(b) Total amounts: 1100.651230 $396,234.00
(c) Interest charges:
1254.681220 $301,123.20
301,123.20 150,000 $151,123.20
(d) The vertical asymptote is x 1000. The closer the payment
is to $1000 per month, the longer the length of the
mortgage will be. Also, the monthly payment must be
greater than $1000.
y ln x 2 78.
0
5
−1
396,234 150,000 $246,234
9
log 2x 3 log 2 9
x 3 9
x 12
log5x 3 log 12
5x 3 12
5x 9
x 9
5
lnx 4 ln 2
x 4 2
x 6
lnx 2 x ln 6
x 2 x 6
x 2 x 6 0
x 3x 2 0
(b)
fx 3 ln x 1
4
−5 10
−6
x 2 or x 3
87.
x
t 12.542 ln , x > 1000
x 1000 88.
ln K
t
0.095
(a) When x $1100.65:
(a)
K 1 2 4 6 8 10 12
1100.65
t 12.542 ln
30 years
1100.65 1000 t 0 7.3 14.6 18.9 21.9 24.2 26.2
The number of years required to multiply the original
investment by K increases with K. However, the larger
the value of K, the fewer the years required to increase
the value of the investment by an additional multiple
of the original investment.
25
20
15
10
5
t
2 4 6 8 10 12
K

89.
98.
ft 80 17 logt 1, 0 ≤ t ≤ 12
(a)
(b)
(c)
fx
(a)
100
0 12
0
f0 80 17 log 1 80.0
f4 80 17 log 5 68.1
(d) f10 80 17 log 11 62.3
ln x
x
x 1 5 10
f x
10 2
10 4
10 6
0 0.322 0.230 0.046 0.00092 0.0000138
Section 3.2 Logarithmic Functions and Their Graphs 279
90.
10 log
(a)
I
1012 10 log 1
10 log 102
10 12 10 log1012 120 decibels
(b)
10
(c) No, the difference is due to the logarithmic
relationship between intensity and number of
decibels.
12 10 log1010 100 decibels
91. False. Reflecting gx about the line y x will determine 92. True, log3 27 3 ⇒ 3
the graph of f x.
3 27.
93.
96.
fx 3 x , gx log 3 x 94.
2
1
−2 −1 1 2
−1
−2
y
f
f and g are inverses. Their graphs
are reflected about the line y x.
g
x
fx 5 x , gx log 5 x 95 .
2
1
−2 −1 1 2
−1
−2
y
f
f and g are inverses. Their graphs
are reflected about the line y x.
fx 10 x , gx log 10 x 97. (a)
2
1
−2 −1 1 2
−1
−2
y
f
g
f and g are inverses. Their graphs are reflected
about the line y x.
x
g
x
(b)
fx ln x,
The natural log function
grows at a slower rate
than the square root
function.
fx ln x,
gx x
gx 4 x
The natural log function
grows at a slower rate than
the fourth root function.
fx e x , gx ln x
2
1
−2 −1 1 2
−1
−2
y
f
f and g are inverses. Their graphs
are reflected about the line y x.
0
0
15
0
0
(b) As x → , fx → 0.
(c)
0.5
0
0
100
40
g
g
f
f
g
x
1000
20,000

280 Chapter 3 Exponential and Logarithmic Functions
99. (a) False. If were an exponential function of then
but not 0. Because one point is
is not an exponential function of
(c) True.
For
y 3, 23 y 1, 2
8
1 y 0, 2
2
0 a 2, x 2
1
y x a
.
y
a
1, 0, y
x.
1 y a a,
x y x,
(b) True. y loga x
,
For a 2, y log2 x.
x 1, log2 1 0
x 2, log2 2 1
x 8, log2 8 3
(d) False. If y were a linear function of x, the slope
between 1, 0 and 2, 1 and the slope between
2, 1 and 8, 3 would be the same. However,
3 1 2 1
and m2
8 2 6 3
Therefore, y is not a linear function of x.
.
m 1 0
1 1
2 1
100. so, for example, if there is no value of y for which 2 If a 1, then every power
of a is equal to 1, so x could only be 1. So, loga x is defined only for 0 < a < 1 and a > 1.
y y loga x ⇒ a a 2,
4.
y x,
102. 101. fx ln x
(a) 4
(b) Increasing on 1,
Decreasing on 0, 1
−1
−2
8
(c) Relative minimum:
1, 0
For Exercises 103–108, use and gx x3 f x 3x 2
1.
103.
105.
107.
f g2 f 2 g2 104.
32 2 2 3 1
8 7
15
fg6 f 6g6 106.
36 26 3 1
20215
4300
f g7 f g7 108.
f 7 3 1
f 342
3342 2
1028
(a) hx lnx2 1
−9
Therefore,
8
−4
f x gx 3x 2 x 3 1
3x 2 x 3 1
3x x 3 3
f g1 31 1 3 3
3 1 3
1.
Therefore, f 3 0 2
0 g 03 1 2.
f x 3x 2
gx x3 1
g f (x g f x g3x 2 3x 2 3 1
Therefore,
g f 3 3 3 2 3 1
7 3 1 344.
9
(b) Increasing on 0,
Decreasing on , 0
(c) Relative minimum:
0, 0

Section 3.3 Properties of Logarithms
■ You should know the following properties of logarithms.
(a)
(b)
(c)
(d) loga un n loga u
loga u
v log loga x
logauv loga u loga v
a u loga v
log10 x
loga x
log10 a
logb x
logb a
■ You should be able to rewrite logarithmic expressions using these properties.
12. log 14 5
14. log 20 0.125
16. log 3 0.015
Section 3.3 Properties of Logarithms 281
log 5 ln 5
log14 ln14 1.161 13. log log 0.4 ln 0.4
90.4 0.417
log 9 ln 9
log 0.125
log 20
log 0.015
log 3
log a x
ln u
lnuv ln u ln v
ln u ln v
v
ln u n n ln u
Vocabulary Check
1. change-of-base 2.
3.
This is the Product Property. Matches (c).
4.
This is the Power Property. Matches (a).
5. loga This is the Quotient Property. Matches (b).
u
v loga u loga v
ln un log x ln x
log a ln a
logauv loga u loga v
n ln u
1. (a)
4. (a)
7. (a)
log 5 x
(b) log 5 x
log 13 x
(b) log 13 x
(b) log 2.6 x
10. log 7 4
log 2.6 x
log x
log 5
ln x
ln 5
log x
log13
ln x
ln13
log x
log 2.6
ln x
ln 2.6
2. (a)
5. (a)
8. (a)
log 3 x
(b) log 3 x
ln x
ln 3
ln x
ln a
log x
log 3
logx 3 log310
10 log x
(b) logx 3 ln310
10 ln x
log 7.1 x
(b) log 7.1 x
log x
log 7.1
ln x
ln 7.1
3. (a)
6. (a)
(b) log 15 x
ln 0.125
ln 20 0.694 15. log log 1250 ln 1250
15 1250 2.633
log 15 ln 15
ln 0.015
ln 3 3.823 17. log4 8 log2 8
log2 4 log2 23 3
log2 22
2
log 15 x
log x
log15
ln x
ln15
logx 3 log34
4 log x
(b) logx 3 ln34
4 ln x
9. log 3 7
log 4 ln 4
log 7 ln 7 0.712 11. log log 4 ln 4
12 4
log12 ln12 2.000
log 7 ln 7
1.771
log 3 ln 3

282 Chapter 3 Exponential and Logarithmic Functions
18.
21.
log 24 2 3 4 log 2 4 2 log 2 3 4 19.
2 log 2 4 4 log 2 3
2 log 2 2 2 4 log 2 3
4 log 2 2 4 log 2 3
4 4 log 2 3
ln5e 6 ln 5 ln e 6 22.
ln 5 6
6 ln 5
log5 1
250 log5 1
125 1
2 20.
log5 1
125 log5 1
2
log 5 5 3 log 5 2 1
3 log 5 2
ln 6 2 ln e
ln 6 2
log 9
3
300 log 100
log 3 log 100
log 3 log 10 2
log 3 2 log 10
log 3 2
ln 6
e 2 ln 6 ln e2 23. log 3 9 2 log 3 3 2
24. log5 1
125 log5 53 3 log5 5 31 3 25. log 4 1
28 4 log2 23 3
4 log2 2 3
41 3
4
26. log6 3 6 log6 613 1
3 log 1
6 6 31 1
3
28.
0.2 log 3 3 4
0.24 0.8
27. log 4 16 1.2 1.2log 4 16 1.2 log 4 4 2 1.22 2.4
log3 810.2 0.2 log3 81 29. log39 is undefined. 9 is not in the domain of log3 x.
30. log216 is undefined because
16 is not in the domain of
log2 x.
31. ln e4.5 4.5 32. 3 ln e
121
12
4 34 ln e
33.
36.
ln 1
ln 1 lne 34.
e
0 1
ln e
2
0 1
2 1
1
2
2 ln e 6 ln e 5 ln e 12 ln e 5
ln e12
e 5
ln e 7
7
ln 4 e 3 ln e 34
3
ln e
4
3
4 1
3
4
37.
log 5 75 log 5 3 log 5 75
3
35. ln e 2 ln e 5 2 5 7
log 5 25
log 5 5 2
2 log 5 5
38. log4 2 log4 32 log4 412 log4 452 39. log4 5x log4 5 log4 x
1
2 log 5
4 4 2 log4 4
1
21 5
21
3
2

53.
55.
ln
3 x
y
1 x
ln
3 y
1
ln x ln y
3
1
3
1
ln x ln y
3
ln x4 y
z 5 ln x4 y ln z 5 56.
ln x 4 ln y ln z 5
4 ln x 1
ln y 5 ln z
2
54.
Section 3.3 Properties of Logarithms 283
40. log3 10z log3 10 log3 z 41. log8 x4 4 log8 x 42. log y
log y log 2
2
43.
57.
59.
log5 5
x log5 5 log5 x 44. log6 z3 3 log6 z 45. lnz ln z12 1
ln z
2
1 log 5 x
46. ln 3 t ln t13 1
3 ln t 47.
49.
51.
ln z 2 lnz 1, z > 1
log 5 x2
y 2 z 3 log 5 x 2 log 5 y 2 z 3 58.
log 5 x 2 log 5 y 2 log 5 z 3
2 log 5 x 2 log 5 y 3 log 5 z
ln 4 x 3 x 2 3 1
4 ln x3 x 2 3 60.
1
4ln x 3 lnx 2 3
1
43 ln x lnx 2 3
3
4 ln x 1 2
4 lnx 3
ln x ln y 2 ln z
ln zz 1 2 ln z lnz 1 2 50.
a 1
log2 log2a 1 log2 9 52.
9
1
2 log 2a 1 log 2 3 2
1
2 log 2a 1 2 log 2 3, a > 1
ln xyz 2 ln x ln y ln z 2 48.
ln x2 1
x 3 lnx2 1 ln x 3
ln x2 x2
ln
y3 y312 lnx 1x 1 ln x 3
lnx 1 lnx 1 3 ln x
6
ln
x2 1 ln 6 lnx2 1
1 x2
ln
2 y3 1
2 ln x2 ln y 3
1
2 ln x 3 ln y
2
ln x 3
ln y
2
x y4
log2 z4 log2 x y4 log2 z4 log 2 x log 2 y 4 log 2 z 4
1
2 log 2 x 4 log 2 y 4 log 2 z
log xy4
z 5 log xy4 log z 5
log x log y 4 log z 5
log x 4 log y 5 log z
lnx 2 x 2 lnx 2 x 2 12
lnxx 2 12
log 4x 2 y log 4 log x 2 log y
ln 6 lnx 2 1 12
ln 6 1
2 lnx2 1
ln x lnx 2 12
ln x 1
2 lnx 2
log 4 2 log x log y

284 Chapter 3 Exponential and Logarithmic Functions
61. ln x ln 3 ln 3x 62. ln y ln t ln yt ln ty 63. log 4 z log 4 y log 4 z
y
64. log 5 8 log 5 t log 5 8
t
65. 2 log 2x 4 log 2x 4 2 66. 2
3 log 7z 2 log 7z 2 23
67. 1
4 log 3 5x log 35x 14 log 3 4 5x 68. 4 log 6 2x log 62x 4 log 6 1
16x 4
69.
71.
ln x 3 lnx 1 ln x lnx 13 70. 2 ln 8 5 lnz 4 ln 82 lnz 45 x
ln
x 13 log x 2 log y 3 log z log x log y 2 log z 3 72.
log x
y2 log z3 log xz3
y2 73. ln x 4lnx 2 lnx 2 ln x 4 lnx 2x 2
74.
75.
76.
ln x 4 lnx 2 4
ln x lnx 2 4 4
x
ln
x2 44 4ln z lnz 5 2 lnz 5 4ln zz 5 lnz 5 2
lnzz 5 4 lnz 5 2
ln z4 z 5 4
z 5 2
1
3 2 lnx 3 ln x lnx2 1 1
3 lnx 32 ln x lnx 2 1
1
3 ln xx 32 lnx 2 1
ln 3
xx 32
x2
1
1 xx 32
ln
3 x2 1
23 ln x lnx 1 lnx 1 2ln x 3 lnx 1 lnx 1
2ln x 3 lnx 1 lnx 1
2ln x 3 lnx 1x 1
x
2 ln
3
x2 1
x
ln
3
x2 1 2
ln 64 lnz 4 5
ln 64z 4 5
3 log 3 x 4 log 3 y 4 log 3 z log 3 x 3 log 3 y 4 log 3 z 4
log 3 x 3 y 4 log 3 z 4
log3 x3y4 z4

77.
78.
80.
82.
1
3 log8 y 2 log8 y 4 log8 y 1 1
log8 y log
3
8 y 42 log8 y 1
log770 1
2 log 1
7 70
2 log7 7 log7 10 81.
10 log
1
2 1 log 7 10
1 1
2 2 log7 10
1
2 log710 by Property 1 and Property 3
I
1012 Difference 10 log
10log3.16 10 7 log1.26 10 5
3.16 107
10log1.26 105 10log2.5079 10 2
10log250.79
24 dB
3.16 105
1.26 107
12 10 log
10
1
3 log 8 y y 42 log 8 y 1
log 8 3 y y 42 log8y 3
y 4
y 1
2 log8 y 1
1
2log4x 1 2 log4x 1 6 log4 x 1
2log4x 1 log4x 12 log4 x6 log 4x 6 x 1x 1
79. log2 The second and third expressions are equal by Property 2.
32
4 log2 32 log2 4 log2 32
log2 4
1
2log 4x 1x 1 2 log 4 x 6
log 4x 1x 1 log 4 x 6
10
12
83.
Section 3.3 Properties of Logarithms 285
10 log
I
1012 10log I log 10 12
10log I 12
120 10 log I
When I 106 :
120 10 log 10 6
120 106
60 decibels
120 10 log2I
120 10log 2 log I
120 10 log I 10 log 2
With both stereos playing, the music is
decibels louder.
10 log 2 3

286 Chapter 3 Exponential and Logarithmic Functions
84.
f t 90 15 logt 1, 0 ≤ t ≤ 12
(a)
(b)
(c)
(d)
(e)
f t 90 logt 1 15
f 0 90
f 4 90 15 log4 1 79.5
f 12 90 15 log12 1 73.3
95
0
70
12
(f) The average score will be 75 when t 9 months. See
graph in (e).
85. By using the regression feature on a graphing calculator we obtain y 256.24 20.8 ln x.
86. (a)
87.
(b)
(d)
80
0
0
See graph in (a).
0.07
0
0
T
30
T 21 54.40.964 t
T 54.40.964 t 21
1
0.0012t 0.016
T 21
1
21
0.0012t 0.016
30
0
0
(e) Since the scatter plot of the original data is so
nicely exponential, there is no need to do the
transformations unless one desires to deal with
smaller numbers. The transformations did not
make the problem simpler.
Taking logs of temperatures led to a linear scatter
plot because the log function increases very slowly
as the x-values
increase. Taking the reciprocals of
the temperatures led to a linear scatter plot because
of the asymptotic nature of the reciprocal function.
80
30
(c)
(g)
75 90 15 logt 1
15 15 logt 1
1 logt 1
10 1 t 1
t (in minutes)
5
0
0
t 9 months
T C
T 21 C
0 78 57 4.043 0.0175
5 66 45 3.807 0.0222
10 57.5 36.5 3.597 0.0274
15 51.2 30.2 3.408 0.0331
20 46.3 25.3 3.231 0.0395
25 42.5 21.5 3.068 0.0465
30 39.6 18.6 2.923 0.0538
lnT 21 0.037t 4
T e 0.037t4 21
This graph is identical to T in (b).
30
lnT 21
1T 21
fx ln x 88. fax fa fx,
a > 0, x > 0
False, f0 0 since 0 is not in the domain of fx.
f1 ln 1 0
True, because fax ln ax ln a ln x fa fx.

89. False. fx f2 ln x ln 2 ln x
2
91. False.
95.
98.
101.
fu 2fv ⇒ ln u 2 ln v ⇒ ln u ln v 2 ⇒ u v 2
93. Let and then bx u and by x logb u y logb v,
v.
u
v
f x log 2 x
−3
bx
bxy
by Then log buv log bb xy x y log b u log b v.
2
3
−3
−1 5
−2
log x ln x
log 2 ln 2
log x ln x
log14 ln14
6
96.
f x log 14 x 99.
f x ln
fx hx by Property 2
x ln x
, gx , hx ln x ln 2
2 ln 2
lnx 2 90. f x false
1
2 fx;
f x log 4 x
2
−1 5
−1
−2
2
−2
Section 3.3 Properties of Logarithms 287
94. Let then and un bnx u b .
x
x logb u,
log b u n log b b nx nx n log b u
log x ln x
log 4 ln 4
log x ln x
log 11.8 ln 11.8
5
97.
f x log 11.8 x 100.
2
1
−1
−2
y
g
f = h
1 2 3 4
can’t be simplified further.
f x lnx ln x12 1 1
ln x
2 2 fx
f x ln x
92. If f x < 0,
then 0 < x < 1.
True
x
f x log 12 x
−3
2
log x ln x
log12 ln12
3
−3
f x log 12.4 x
−1 5
−2
log x ln x
log 12.4 ln 12.4
6

288 Chapter 3 Exponential and Logarithmic Functions
102.
ln 2 0.6931, ln 3 1.0986, ln 5 1.6094
ln 2 0.6931
ln 3 1.0986
ln 4 ln2 2 ln 2 ln 2 0.6931 0.6931 1.3862
ln 5 1.6094
ln 6 ln2 3 ln 2 ln 3 0.6931 1.0986 1.7917
ln 8 ln 2 3 3 ln 2 30.6931 2.0793
ln 9 ln 3 2 2 ln 3 21.0986 2.1972
ln 10 ln5 2 ln 5 ln 2 1.6094 0.6931 2.3025
ln 12 ln2 2 3 ln 2 2 ln 3 2 ln 2 ln 3 20.6931 1.0986 2.4848
ln 15 ln5 3 ln 5 ln 3 1.6094 1.0986 2.7080
ln 16 ln 2 4 4 ln 2 40.6931 2.7724
ln 18 ln3 2 2 ln 3 2 ln 2 2 ln 3 ln 2 21.0986 0.6931 2.8903
ln 20 ln5 2 2 ln 5 ln 2 2 ln 5 2 ln 2 1.6094 20.6931 2.9956
103. 24xy2
16x 3 y
24xx3 3x4
16yy2
2y3, x 0 104. 2x2
3y 3
3y
2x23 105. 18x 3 y 4 3 18x 3 y 4 3 18x3 y 4 3
107.
109.
18x3y4 1 if x 0, y 0. 106.
3 3x 2 2x 1 0 108.
3x 1x 1 0
3x 1 0 ⇒ x 1
3
x 1 0 ⇒ x 1
2 x
3x 1 4
3x 1x 24
3x 2 x 8 0
x 1 ± 12 438
23
1 ± 97
6
110.
xyx 1 y 1 1
4x 2 5x 1 0
4x 1x 1 0
4x 1 0 ⇒ x 1
4
x 1 0 ⇒ x 1
The zeros are x 1
4 , 1.
The zeros are
1 ±31
.
2
3y3
2x2 27y3
3 8x6 xy
x 1 y 1
xy
1x 1y
xy xy2
y xxy x y
2 ±22 4215
22
5 2x
x 1 3
53 2xx 1
15 2x 2 2x
x
2 ±124
x
4
1 ±31
x
2
0 2x 2 2x 15

Section 3.4 Exponential and Logarithmic Equations
Vocabulary Check
Section 3.4 Exponential and Logarithmic Equations 289
■ To solve an exponential equation, isolate the exponential expression, then take the logarithm of both sides.
Then solve for the variable.
1. 2.
■ To solve a logarithmic equation, rewrite it in exponential form. Then solve for the variable.
1. 2.
■ If and we have the following:
1.
2. ax ay e
a > 0 a 1
loga x loga y ⇔ x y
⇔ x y
ln x a x
log ln e
a x x
x loga a x
x x
■ Check for extraneous solutions.
1. solve 2. (a) x y (b) x y
3. extraneous
(c) x (d) x
1. 4
(a) x 5
2x7 64 2. 2
(a) x 1
3x1 32
(b)
(b)
(c)
4 257 4 3 64
Yes, x 5 is a solution.
x 2
4 227 43 1
64 64
No, x 2 is not a solution.
3.
(a)
3e 2e252 3ee25 x 2 e
75
25
3ex2 75 4. 2e
(a)
5x2 12
No, x 2 e is not a solution.
25
x 2 ln 25
3e 2ln 252 3e ln 25 325 75
Yes, x 2 ln 25 is a solution.
x 1.219
3e 1.2192 3e 3.219 75
Yes, x 1.219 is a solution.
(b)
(b)
2 311 2 2 1
4
No, x 1 is not a solution.
x 2
2 321 2 7 128
No, x 2 is not a solution.
2e 5152ln 62 2e2ln 62
Yes, x is a solution.
1
2 ln 6
5
x
x 1
2 ln 6
5
2e ln 6 2 6 12
ln 6
5 ln 2
2e 5[ln 65 ln 22 ln 6ln 22
2e
2e
2 97.9995 195.999
ln 6
No, x is not a solution.
5 ln 2
(c)
x 0.0416
2.5852
2e 50.04162 2e 1.792 26.00144 12
Yes, x 0.0416 is an approximate solution.

290 Chapter 3 Exponential and Logarithmic Functions
5. log43x 3 ⇒ 3x 4
(a) x 21.333
3 ⇒ 3x 64 6. log2x 3 10
(a)
x 1021
(b)
(c)
321.333 64
Yes, 21.333 is an approximate solution.
x 4
34 12 64
No, x 4 is not a solution.
x 64
3
3 64
3 64
Yes, x is a solution.
64
3
(b)
(c)
log 21021 3 log 21024
Since 2 x 1021 is a solution.
10 1024,
x 17
log 217 3 log 220
Since 2 x 17 is not a solution.
10 20,
x 10 2 3 97
log 297 3 log 2100
Since 10 is not a solution.
2 2 3
10 100,
7.
(a)
No, is not a solution.
(b)
Yes, x is a solution.
(c)
x 163.650
1
23 e5.8 ln2
1
23 e5.8 3 lne5.8 x
5.8
1
23 e5.8 x
1
ln2
23 ln 5.8
1
x
23 ln 5.8 3 lnln 5.8 5.8
1
ln2x 3 5.8 8.
23 ln 5.8
(a)
Yes, x 1 e is a solution.
(b)
x 45.701
ln45.701 1 ln44.701 3.8
Yes, x 45.701 is an approximate solution.
(c)
x 1 ln 3.8
3.8
ln1 e3.8 1 ln e3.8 x 1 e
3.8
3.8
lnx 1 3.8
9.
13.
17.
4 x 4 2
x 2
ln2163.650 3 ln 330.3 5.8
Yes, x 163.650 is an approximate solution.
4 x 16 10.
ln x ln 2 0 14.
ln x ln 2
x 2
ln x 1 18.
e ln x e 1
x e 1
x 0.368
3 x 243 11.
3 x 3 5
x 5
ln x ln 5 0 15.
e ln x e 7
x e 7
ln x ln 5
x 5
x 0.000912
x 5
ln1 ln 3.8 1 lnln 3.8 0.289
No, x 1 ln 3.8 is not a solution.
1
2 x
32 12.
2 x 2 5
x 5
e x 2 16.
ln e x ln 2
x ln 2
x 0.693
1
4 x
64
4 x 4 3
x 3
x 3
e x 4
ln e x ln 4
x ln 4
x 1.386
ln x 7 19. log4 x 3 20. log5 x 3
4 log 4 x 4 3
x 4 3
x 64
x 5 3
x 1
125
or 0.008

21.
25.
28.
31.
34.
e x2
x 2 x 2 2x
2x 2 2x 0
2xx 1 0
e x2 2x 29.
x 0, x 1
2e x 10 32.
e x 5
ln e x ln 5
x ln 5 1.609
6 x 10 47 35.
6 x 37
x log 6 37
x
ln 37
ln 6
x 2.015
37. 5t2 0.20 38.
40.
fx gx 22.
2 x 8
2 x 2 3
x 3
Point of intersection:
3, 8
5 t2 1
5
5 t2 5 1
t
2 1
t 2
27 x 9
27 x 27 23
x 2
3
Point of intersection:
2
3, 9
e x e x2 2 26.
x x 2 2
0 x 2 x 2
0 x 1x 2
x 1 or x 2
2 x3 32
x 3 log 2 32
x 3 5
x 8
fx gx 23.
x 2 2x 8 0
x 4x 2 0
3 x 5
log 3 3 x log 3 5
Section 3.4 Exponential and Logarithmic Equations 291
43 x 20 30.
e x 91
4
x log 3 5
x 1.465
ln e x ln 91
4
x ln 91
4
3.125
log 5
log 3
or ln 5
ln 3
4e x 91 33.
3 2x 80 36.
ln 3 2x ln 80
2x ln 3 ln 80
x
3t
ln 80
1.994
2 ln 3
4 3t 0.10 39.
ln 4 3t ln 0.10
3t ln 4 ln 0.10
ln 0.10
ln 4
ln 0.10
t 0.554
3 ln 4
fx gx 24.
log 3 x 2
x 3 2
x 9
Point of intersection:
9, 2
e 2 x e x2 8 27.
2x x 2 8
x 2, x 4
25 x 32
5 x 16
x log 516
x
ln 16
ln 5
x 1.723
e x 9 19
e x 28
ln e x ln 28
x ln 28 3.332
6 5x 3000
ln 6 5x ln 3000
5x ln 6 ln 3000
5x
x
3 x1 27
3 x1 3 3
x 1 3
x 4
fx gx
lnx 4 0
x 2 x 1 0
e lnx4 e 0
x 4 1
Point of intersection: 5, 0
e x2 3 e x2
ln 3000
ln 6
ln 3000
5 ln 6
x 5
x 2 3 x 2
By the Quadratic Formula
x 1.618 or x 0.618.
0.894

292 Chapter 3 Exponential and Logarithmic Functions
41.
43.
46.
49.
2 3x 565 42.
ln 2 3x ln 565
3 x ln 2 ln 565
3 ln 2 x ln 2 ln 565
x ln 2 ln 565 3 ln 2
x ln 2 3 ln 2 ln 565
x
3 ln 2 ln 565
ln 2
3
ln 565
ln 2
6.142
810 3x 12 44.
10 3x 12
8
log 10 3x log 3
2
3x log 3
2
x 1 3
log
3 2
0.059
83 6x 40 47.
3 6x 5
ln 3 6x ln 5
6 x ln 3 ln 5
6 x
x
ln 5
ln 3
ln 5
6
ln 3
x 6
ln 5
4.535
ln 3
500e x 300 50.
e x 3
5
x ln 3
5
x ln 3
5
ln 5
3 0.511
10 x6 7
5
log 10 x6 log 7
5
x 6 log 7
5
x 6 log 7
5
6.146
8 2x 431
ln 8 2x ln 431
2 x ln 8 ln 431
2 ln 8 x ln 8 ln 431
x ln 8 ln 431 ln 8 2
x ln 8 ln 431 ln 64
x
5 10 x6 7 45.
e 3x 12 48.
3x ln 12
x
ln 12
3
0.828
1000e 4x 75 51.
e 4x 3
40
ln e 4x ln 3
40
4x ln 3
40
x 1 3
4 ln 40
0.648
2e x 2
e x 1
x ln 1 0
ln 431 ln 64
ln 8
35 x1 21
5 x1 7
ln 5 x1 ln 7
x 1 ln 5 ln 7
x 1
e 2x 50
ln e 2x ln 50
2x ln 50
x
4.917
ln 7
ln 5
x 1
ln 50
2
1.956
ln 7
2.209
ln 5
7 2ex 5 52. 14 3ex 11
3e x 25
e x 25
3
ln e x ln 25
3
x ln 25
3
2.120

53.
55.
57.
59.
62 3x1 7 9
6 2 3x1 16
2 3x1 8
3
log 2 2 3x1 log 2 8
3
8 log83
3x 1 log2 3 log 2
e 2x 4e x 5 0 56.
e x 1e x 5 0
or e
(No solution) x ln 5 1.609
x e 5
x 1
e 2x 3e x 4 0
e x 1e x 4 0
e x 10 ⇒ e x 1
e x > 0
Not possible since for all x.
or ln83
ln 2
x 1
3 log83
1 0.805
log 2
e x 40 ⇒ e x 4 ⇒ x ln 4 1.386
500
100 e x2 20 60.
500 20100 e x2
25 100 e x2
e x2 75
x
ln 75
2
x 2 ln 75 8.635
Section 3.4 Exponential and Logarithmic Equations 293
54.
58.
400 3501 e x
8
1 ex
7
8
1 ex
7
1
ex
7
ln 1
ln ex
7
x ln 1
7
x ln 7 1
x ln 7
x ln 7 1.946
84 62x 13 41
84 62x 28
4 62x 3.5
6 2x log 4 3.5
6 2x
ln 3.5
ln 4
2x 6
x 3
e 2x 5e x 6 0
e x 2e x 3 0
ln 3.5
ln 4
ln 3.5
2.548
2 ln 4
e x 2 or e x 3
x ln 2 0.693 or x ln 3 1.099
e 2x 9e x 36 0
e x 2 9e x 36 0
Because the discriminant is 9 there
is no solution.
2 4136 63,
400
350 61.
1 ex 3000
2 e2x 2
3000 22 e 2x
1500 2 e 2x
1498 e 2x
ln 1498 2x
x
ln 1498
2
3.656

294 Chapter 3 Exponential and Logarithmic Functions
62.
64.
66.
68.
119
e6x 7
14
119 7e 6x 14
17 e 6x 14
31 e 6x
ln 31 ln e 6x
ln 31 6x
x
ln 31
6
2.471
4
40 9t
21 65.
3.938225 9t 21
ln 3.938225 9t ln 21
9t ln 3.938225 ln 21
t
ln 21
0.247
9 ln 3.938225
0.878
16
26 3t
30 67.
0.878
ln16
26 3t
ln 30
0.878
3t ln16
26 ln 30
t
1 ln 3.75 x
1 ln 3.75 x
2.322 x
The zero is 2.322.
0.572
f x 4e x1 15
ln 30
3 ln16 0.878 0.409
26
0 4e x1 15 −5 5
15 4e x1
3.75 e x1
ln 3.75 x 1
20
−20
63.
0.065
1
365 365t
4
0.065
ln1
365 365t
ln 4
0.065
365t ln1
365 ln 4
t
t
0.10
1
12 12t
2
0.10
ln1
12 12t
ln 2
0.10
12t ln1
12 ln 2
gx 6e 1x 25
Algebraically:
6e 1x 25
e 1x 25
6
1 x ln 25
6
x 1 ln 25
6
x 0.427
The zero is x 0.427.
69. f x 3e3x2 962
Algebraically:
3e 3x2 962
e 3x2 962
3
3x
2
ln 962
3
x 2 962
ln 3 3
x 3.847
The zero is x 3.847.
ln 4
365 ln1 0.065 21.330
365
ln 2
12 ln1 0.10 6.960
12
6
−6 15
−30
300
−6 9
−1200

70.
72.
75.
79.
83.
gx 8e 5
2x3 11 71.
8e 2x3 11 −3 7
e 2x3 1.375
2x
3
ln 1.375
x 1.5 ln 1.375
x 0.478
The zero is 0.478.
f x e 1.8x 7 73.
e 1.8x 7 0
e 1.8x 7
e 1.8x 7
1.8x ln 7
x
x 1.081
The zero is 1.081.
13
−5 5
−7
ln 7
1.8
ln x 3 76.
x e 3 0.050
log x 6 80.
x 10 6
1,000,000.000
ln x 2 1 84.
x 2 e 1
x 2 e 2
x e 2 2
5.389
−15
e ln x e 2
x e 2 7.389
Algebraically:
e 0.125t 80
e 0.125t 8
0.125t ln 8
t
Section 3.4 Exponential and Logarithmic Equations 295
ht e 0.125t 8 74.
t 16.636
The zero is t 16.636.
−40 40
2
−10
ln 8
0.125
gt e 0.09t 3
Algebraically:
e 0.09t 3 −20 40
0.09t ln 3
t
ln 3
0.09
t 12.207
The zero is t 12.207.
f x e 2.724x 29
e 2.724x 29
2.724x ln 29
x
ln 29
2.724
x 1.236
The zero is 1.236.
10
−5 5
ln x 2 77. ln 2x 2.4 78. ln 4x 1
10 log 3z 10 2
3z 100
z 100
3
33.333
2x e 2.4
x e2.4
2
5.512
−35
8
−4
e ln 4x e 1
4x e
log 3z 2 81. 3 ln 5x 10 82. 2 ln x 7
lnx 8 5 85.
e lnx8 e 5
x 8 e 5
x 8 e 10
x e 10 8
22,034.466
ln 5x 10
3
5x e 103
x e103
5
3 ln x 2
ln x 2
3
x e 23
0.513
5.606
x e
0.680
4
ln x 7
2
e ln x e 72
x e 72 33.115
7 3 ln x 5 86. 2 6 ln x 10
6 ln x 8
ln x 4
3
e ln x e 43
x e 43
0.264

296 Chapter 3 Exponential and Logarithmic Functions
87.
6 log 30.5x 11 88.
log 30.5x 11
6
3 log 30.5x 3 116
0.5x 3 116
x 23 116 14.988
x e2
1 e2 1.157
This negative value is extraneous. The equation has
no solution.
5 log 10x 2 11
log 10x 2 11
5
10 log 10x2 10 115
x 2 10 115
x 10 115 2 160.489
89.
x1 e2 e2 x e2x e2 x e2x e2 x e2x 1
x
ln x lnx 1 2 90.
x
ln 2
x 1
e2
x 1
x
1 ± 1 4e
x
2
1 1 4e
The only solution is x 1.223.
2
2 xx 1 e
x e 0
1
elnxx1 e1 ln x lnx 1 1
lnxx 1 1
91.
93.
ln x lnx 2 1 92.
lnxx 2 1
xx 2 e 1
x 2 2x e 0
x
2 ± 4 4e
2
2 ± 21 e
2
1 ± 1 e
The negative value is extraneous. The only solution is
x 1 1 e 2.928.
ln x 5 lnx 1 lnx 1 94.
x 1
lnx 5 lnx 1
x 5
x 2 5x 6 0
x 2x 3 0
x 1
x 1
x 5x 1 x 1
x 2 6x 5 x 1
x 2 or x 3
Both of these solutions are extraneous, so the equation
has no solution.
95. log22x 3 log2x 4
2x 3 x 4
x 7
ln x lnx 3 1
lnxx 3 1
e lnxx3 e 1
xx 3 e 1
x 2 3x e 0
x
3 ± 9 4e
2
3 9 4e
The only solution is x 0.729.
2
lnx 1 lnx 2 ln x
x 1
ln ln x
x 2
x 1
x
x 2
3 ±32 411
21
x 1 x 2 2x
x
3 ±13
x
2
0 x 2 3x 1
3.303 x
(The negative apparent solution is extraneous.)

96.
97.
0 x
1 ± 17
x
2
Quadratic Formula
Choosing the positive value of x (the negative value is
extraneous), we have
2 x 4 x
x 4
2 logx 4 log x logx 2
x 4
log
x logx 2
x 4
x 2
x
2x
98.
99.
logx 6 log2x 1
x
x 6 2x 1
7 x
1 17
2
1.562.
x
4
412
x 1 log4xx1 412 x 2x 1
x 2x 2
x 2
x 2
Section 3.4 Exponential and Logarithmic Equations 297
The apparent solution x 7 is extraneous, because the domain of the logarithm function is positive numbers,
and 7 6 and 27 1 are negative. There is no solution.
log4 x log4x 1
x 1
log4
x 1 2
1
2
101. log 8x log1 x 2
8x
log 2
1 x
The only solution is x
8x 1001 x
8x
102
1 x
4x 2 100x 625 625x
4x 2 725x 625 0
2x 251 x 25 25x
2x 25 25x
2x 25 2 25x 2
x 725 ± 7252 44625
24
2529 533
8
180.384.
x 0.866 (extraneous) or x 180.384
100.
725 ± 515,625
8
log 2 x log 2x 2 log 2x 6
log 2xx 2 log 2x 6
xx 2 x 6
x 2 x 6 0
x 3x 2 0
x 3 or x 2
The value x 3 is extraneous. The only solution is
x 2.
log 3 x log 3x 8 2
log 3xx 8 2
3 log 3x 2 8x 3 2
x 2 8x 9
x 2 8x 9 0
x 9x 1 0
x 9 or x 1
The value x 1 is extraneous. The only solution is
x 9.
2529 ± 533
8

298 Chapter 3 Exponential and Logarithmic Functions
102.
103.
105.
log 4x log12 x 2
4x
log 2
12 x
10 log4x(12x 10 2
4x
100
12 x
4x 1200 100x
x 2 600x 90,000 625x
x 2 1225x 90,000 0
The only solution is x
y 2 2 x
From the graph we have
x 2.807 when y 7.
Algebraically:
2 x 7
ln 2 x ln 7
x ln 2 ln 7
x
4x 10012 x
4x 1200 100x
x 300 25x
x 300 2 25x 2
ln 7
2.807
ln 2
x 1225 ± 12252 4190,000
2
x
x
1225 ± 1,140,625
2
1225 ± 12573
2
x 78.500 extraneous or x 1146.500
1225 12573
2
1146.500.
y 10
1 7 104.
−8 10
y 5
1 3 106.
y 2 ln x
From the graph we have
x 20.086 when y 3.
Algebraically:
3 ln x 0
ln x 3
x e 3 20.086
−5
−1
−2
30
500 1500e x2
1
ex2
3
ln 1
3 x
2
2 ln 1
x
3
2.197 x
The solution is x 2.197.
10 4 lnx 2 0
4 lnx 2 10
lnx 2 2.5
e lnx2 e 2.5
x 2 e 2.5
x e 2.5 2
x 14.182
The solution is x 14.182.
−2
−5
800
−200
18
−3
10
30

107. (a)
109.
A Pe rt (b)
5000 2500e 0.085t
2 e 0.085t
ln 2 0.085t
ln 2
t
0.085
t 8.2 years
p 500 0.5e 0.004x
(a)
p 350
350 500 0.5e 0.004x
300 e 0.004x
0.004x ln 300
x 1426 units
4
110. p 50001
4 e
(a) When p $600:
0.002x
0.12 1
4
0.88
4 e0.002x 0.48 0.88e 0.002x
6
e0.002x
11
ln 6
ln e0.002x
11
ln 6
11 0.002x
x ln611
0.002
111. V 6.7e
(a) 10
48.1t , t ≥ 0
0
0
A Pe rt 108. (a)
7500 2500e 0.085t
3 e 0.085t
ln 3 0.085t
ln 3
t
0.085
(b)
4
600 50001
4 e0.002x 4
4 e 0.002x
4 3.52 0.88e 0.002x
1500
303 units
t 12.9 years
p 300
400 e 0.004x
0.004x ln 400
Section 3.4 Exponential and Logarithmic Equations 299
300 500 0.5e 0.004x
x 1498 units
r 0.12 (b)
A Pe rt
5000 2500e 0.12t
2 e 0.12t
ln 2 ln e 0.12t
ln 2 0.12t
ln 2
t
0.12
t 5.8 years
(b) When p $400:
0.08 1
4
0.92
4 e0.002x 4 3.68 0.92e 0.002x
0.32 0.92e 0.002x
8
e0.002x
23
ln 8
ln e0.002x
23
ln 8
23 0.002x
x ln823
0.002
(b) As t → , V → 6.7. (c)
Horizontal asymptote: V 6.7
The yield will approach
6.7 million cubic feet per acre.
4
4 e 0.002x
528 units
r 0.12
A Pe rt
7500 2500e 0.12t
3 e 0.12t
ln 3 ln e 0.12t
ln 3 0.12t
ln 3
t
0.12
4
400 50001
4 e0.002x 1.3 6.7e 48.1t
1.3
e48.1t
6.7
ln 13 48.1
67 t
t 9.2 years
t 48.1
29.3 years
ln1367

300 Chapter 3 Exponential and Logarithmic Functions
112. N 6810
When N 21:
0.04x 113. y 7312 630.0 ln t, 5 ≤ t ≤ 12
7312 630.0 ln t 5800
114.
21 6810 0.04x
21
100.04x
68
log 10 21
68 0.04x
x log 102168
0.04
9000 4381 1883.6 ln t
4619 1883.6 ln t
ln t 4619
2.45222
1883.6
t e 2.45222 11.6
12.76 inches
y 4381 1883.6 ln t, 5 ≤ t ≤ 13
630.0 ln t 1512
ln t 2.4
t e 2.4 11
t 11 corresponds to the year 2001.
Since t 5 represents 1995, t 11.6 indicates that the number of daily fee golf facilities in the U.S.
reached 9000 in 2001.
115. (a) From the graph shown in the textbook, we see horizontal asymptotes at y 0 and y 100.
These represent the lower and upper percent bounds; the range falls between 0% and 100%.
(b) Males
50
1 e 0.6114x69.71 2
e 0.6114x69.71 1
0.6114x 69.71 ln 1
0.6114x 69.71 0
116. P
(a) 1.0
0.83
1 e0.2n 0 40
0
100
1 e 0.6114x69.71
x 69.71 inches
(b) Horizontal asymptotes: P 0, P 0.83
The upper asymptote, P 0.83, indicates that the
proportion of correct responses will approach 0.83
as the number of trials increases.
Females
50
1 e 0.66607x64.51 2
e 0.6667x64.51 1
0.66607x 64.51 ln 1
0.66607x 64.51 0
100
1 e 0.66607x64.51
x 64.51 inches
(c) When P 60% or P 0.60:
0.60
1 e 0.2n 0.83
0.60
0.83
1 e 0.2n
e0.2n 0.83
1
0.60
ln e0.2n ln 0.83
1 0.60
ln
n
0.83
0.2n ln
0.60
0.2
0.83
1 0.60
1
5 trials

117. y 3.00 11.88 ln x
(a)
36.94
x
x 0.2 0.4 0.6 0.8 1.0
(b)
200
0
0
y 162.6 78.5 52.5 40.5 33.9
The model seems to fit the data well.
(c) When y 30:
1.2
30 3.00 11.88 ln x 36.94
x
Add the graph of y 30 to the graph in part (a) and
estimate the point of intersection of the two graphs.
We find that x 1.20 meters.
(d) No, it is probably not practical to lower the number
of gs experienced during impact to less than 23
because the required distance traveled at y 23 is
x 2.27 meters. It is probably not practical to
design a car allowing a passenger to move forward
2.27 meters (or 7.45 feet) during an impact.
True by Property 1 in Section 3.3.
Section 3.4 Exponential and Logarithmic Equations 301
119. logauv loga u loga v 120. logau v loga uloga v
121.
log au v log a u log a v 122.
False.
1.95 log100 10
log 100 log 10 1
False.
2.04 log 1010 100 log 10 10log 10 100 2
u
logav True by Property 2 in Section 3.3.
loga u loga v 123. Yes, a logarithmic equation can
have more than one extraneous
solution. See Exercise 93.
124.
(a) A 2Pe This doubles your money.
(b)
rt 2Pert A Pe
rt 125. Yes.
Time to Double Time to Quadruple
4P Pert 2P Pert (c)
A Pe 2rt Pe rt e rt e rt Pe rt
A Pe r2t Pe rt e rt e rt Pe rt
Doubling the interest rate yields the same result as
doubling the number of years.
If 2 > e (i.e., rt < ln 2),
then doubling your
investment would yield the most money. If
rt > ln 2, then doubling either the interest rate
or the number of years would yield more money.
rt
118. T 201 72h (a) From the graph in the textbook we see a horizontal
asymptote at T 20. This represents the room
temperature.
(b)
ln47
ln 2
2 e rt
ln 2 rt
ln 2
t
r
100 201 72 h
ln 4
7
5 1 72 h
4 72 h
4
2h
7
ln 2h
ln 4
h ln 2
7
h
h 0.81 hour
4 e rt
ln 4 rt
2 ln 2
t
r
Thus, the time to quadruple is twice as long as the
time to double.

302 Chapter 3 Exponential and Logarithmic Functions
126. (a) When solving an exponential equation, rewrite the
original equation in a form that allows you to use the
One-to-One Property if and only if or
rewrite the original equation in logarithmic form and
use the Inverse Property loga ax a x y
x.
x ay 127.
130.
132.
134.
48x 2 y 5 16x 2 y 4 3y 128.
4 x y2 3y
3
10 2
310 2
10 4
310 2
6
3 10 2
10 2 10 2
10 2
2
1
10 1
2
8
6
4
2
y
−6 −4 −2
−2
−4
−6
2 4 6 8
6
4
1
y
−6 −4 −2
−2
−6
2
4
6
x
x
(b) When solving a logarithmic equation, rewrite the
original equation in a form that allows you to use the
One-to-One Property if and only if
or rewrite the original equation in exponential
form and use the Inverse Property aloga x loga x loga y
x y
x.
32 225 16 2 25 129.
42 10
131. f x x 9
Domain: all real numbers x
133.
y- intercept: 0, 9
y-axis
symmetry
x
y
0
±1
±2
9 10 11 12
gx 2x,
x 2 4,
325 3 15 3 375
±3
x < 0
x ≥ 0
Domain: all real numbers xx-
intercept: 2, 0
y-intercept:
0, 4
135. log6 9 log10 9 ln 9
log10 6 ln 6 1.226 136. log3 4 log10 4 ln 4
1.262
log10 3 ln 3
137. log34 5 log10 5 ln 5
log1034 ln34 5.595 138. log8 22 log10 22 ln 22
1.486
log10 8 ln 8
x
y
3
6
2
4
1
2
0.5
1
3 125 3 5 3 3
14
12
8
6
4
2
−8 −6 −4 −2 2 4 6 8
−2
5
4
3
2
1
−4 −3 −2 −1 1 3 4
−3
y
y
0 1 2 3
4 3 2 5
x
x

Section 3.5 Exponential and Logarithmic Models
■ You should be able to solve growth and decay problems.
(a) Exponential growth if
(b) Exponential decay if b > 0 and y ae
■ You should be able to use the Gaussian model
bx b > 0 and y ae
.
bx .
y ae xb2 c .
■ You should be able to use the logistic growth model
a
y .
1 berx ■ You should be able to use the logarithmic models
y a b ln x, y a b log x.
Vocabulary Check
1.
4.
Section 3.5 Exponential and Logarithmic Models 303
1. y ae 2. y a b ln x; y a b log x 3. normally distributed
bx ; y aebx 4. bell; average value 5. sigmoidal
y 2e x4 2.
This is an exponential growth
model. Matches graph (c).
y 3e x22 5 5.
This is a Gaussian model.
Matches graph (a).
y 6e x4 3.
This is an exponential decay
model. Matches graph (e).
y lnx 1 6.
This is a logarithmic model shifted
left one unit. Matches graph (d).
y 6 logx 2
This is a logarithmic function
shifted up six units and left two
units. Matches graph (b).
4
y
1 e
This is a logistic growth model.
Matches graph (f).
2x
7. Since the time to double is given by
2000 1000e and we have
0.035t
A 1000e0.035t , 8. Since the time to double is given by
1500 750e and we have
0.105t A 750e
,
0.105t ,
2 e 0.035t
ln 2 ln e 0.035t
ln 2 0.035t
t
ln 2
19.8 years.
0.035
Amount after 10 years: A 1000e 0.35 $1419.07
1500 750e 0.105t
2 e 0.105t
ln 2 ln e 0.105t
ln 2 0.105t
t
ln 2
6.60 years.
0.105
Amount after 10 years: A 750e 0.10510 $2143.24

304 Chapter 3 Exponential and Logarithmic Functions
9. Since A 750e when t 7.75, we have
the following.
rt and A 1500 10. Since A 10,000e and A 20,000 when t 12,
we have
rt
1500 750e 7.75r
2 e 7.75r
ln 2 ln e 7.75r
ln 2 7.75r
r
ln 2
0.089438 8.9438%
7.75
Amount after 10 years: A 750e 0.08943810 $1834.37
The time to double is given by
1000 500e 0.110t
t
r ln1505.00500
10
ln 2
6.3 years.
0.110
20,000 10,000e 12r
2 e 12r
ln 2 ln e 12r
ln 2 12r
r
ln 2
12
Amount after 10 years:
0.057762 5.7762%.
A 10,000e 0.05776210 $17,817.97
11. Since when
we have the following.
1505.00 500e
0.110 11.0%
10r
A 500e
t 10,
rt and A $1505.00 12. Since and when we have
19,205
19,205 600e
e10r
600 10r
A 600e A 19,205 t 10,
rt
ln 19,205
600 ln e10r
ln 19,205
600 10r
The time to double is given by
1200 600e 0.3466t
t
r ln19,205600
10
ln 2
2 years.
0.3466
0.3466 or 34.66%.
13. Since and when
we have the following.
10,000.00 Pe
10,000.00
P $6376.28
e0.04510 ln 2
The time to double is given by t 15.40 years.
0.045 0.04510
A Pe A 10,000.00 t 10,
0.045t 14. Since and when we have
P
ln 2
The time to double is given by t 34.7 years.
0.02 2000
2000 Pe
e0.0210 $1637.46.
0.0210
A Pe A 2000 t 10,
0.02t
15.
0.075
500,000 P1
12 1220
P
500,000
0.075
1
12 1220
500,000
$112,087.09
240 1.00625
16.
r
A P1 n nt
0.12
500,000 P1
12 12(40)
P $4214.16

17.
P 1000, r 11%
(a)
(c)
n 1
1 0.11 t 2
t ln 1.11 ln 2
t
n 365
0.11
1
365 365t
2
ln 2
6.642 years
ln 1.11
0.11
365t ln1 ln 2
365
t
18. P 1000, r 10.5% 0.105
19.
20.
21.
(a) n 1
(c)
t
n 365
t
3P Pe rt
ln 3 rt
ln 3
t
r
60
0
0
3 e rt
ln 2
6.94 years
ln1 0.105
ln 2
365 ln1 0.105 6.602 years
365
0.16
ln 2
365 ln1 0.11 6.302 years
365
Section 3.5 Exponential and Logarithmic Models 305
(b)
0.11
1
12 12t
2
0.11
12t ln1
12 ln 2
(d) Compounded continuously
e 0.11t 2
0.11t ln 2
t
(b) n 12
t
n 12
t
ln 2
6.301 years
0.11
(d) Compounded continuously
t
ln 2
12 ln1 0.105 6.63 years
12
ln 2
6.601 years
0.105
r 2% 4% 6% 8% 10% 12%
t
ln 3
r
(years) 54.93 27.47 18.31 13.73 10.99 9.16
Using the power regression feature of a graphing utility, t 1.099r 1 .
3P P1 r t
ln 3 ln1 r t
ln 3 t ln1 r
ln 3
t
ln1 r
3 1 r t
r 2% 4% 6% 8% 10% 12%
t
ln 3
ln1 r
(years) 55.48 28.01 18.85 14.27 11.53 9.69
ln 2
12 ln1 0.11 6.330 years
12

306 Chapter 3 Exponential and Logarithmic Functions
22.
24.
27.
60 23. Continuous compounding results in faster growth.
0
0
0.16
Using the power regression feature of a graphing utility,
t 1.222r 1 .
2
A = 1 + 0.055 ( 365 ) 365t
[[[[
0 10
0
A = 1 + 0.06[[[[ t
From the graph, 2% compounded
daily grows faster than 6% simple
interest.
5 1
25.
1
2 C Cek5715 28.
0.5 e k5715
ln 0.5 ln e k5715
ln 0.5 k5715
k
ln 0.5
5715
Given y 2 grams after 1000
years, we have
2 Celn 0.557151000
C 2.26 grams.
0.5 e k1599
ln 0.5 ln e k1599
ln 0.5 k1599
k
ln 0.5
1599
A 1 0.075t and A e 0.07t
Amount (in dollars)
Given C 10 grams after
1000 years, we have
y 10eln 0.515991000
6.48 grams.
2.00
1.75
1.50
1.25
1.00
A
A = e0.07t
A = 1 + 0.075[[[[ t
2 4 6 8 10
Time (in years)
1
2 C Cek1599 26.
1
2 C Cek5715 29.
1
ek5715
2
ln 1
ln ek5715
2
ln 1
k5715
2
k ln12
5715
Given C 3 grams, after 1000
years we have
y 3e ln1257151000
y 2.66 grams.
t
1
C Cek1599
2
1
ek1599
2
ln 1
ln ek1599
2
ln 1
k1599
2
k ln12
1599
Given y 1.5 grams after 1000
years, we have
1.5 Ce ln1215991000
C 2.31 grams.
1
C Cek24,100
2
0.5 e k24,100
ln 0.5 ln e k24,100
ln 0.5 k24,100
k
ln 0.5
24,100
Given y 2.1 grams after 1000
years, we have
2.1 Celn 0.524,1001000
C 2.16 grams.

30.
33.
35.
36.
1
2 C Cek24,100 31.
1
ek24,100
2
ln 1
ln ek24,100
2
ln 1
k24,100
2
k ln12
24,100
Given y 0.4 grams after 1000
years, we have
0.4 Ce ln1224,1001000
C 0.41 grams.
ln15
4
—CONTINUED—
10 e b3
ln 10 3b
Thus, y e 0.7675x .
Section 3.5 Exponential and Logarithmic Models 307
y ae bx 32.
1 ae b0 ⇒ 1 a
ln 10
b ⇒ b 0.7675
3
y ae bx 34.
5 ae b0 ⇒ 5 a
1 5e b4
1
e4b
5
ln 1
4b 5
b ⇒ b 0.4024
Thus, y 5e 0.4024x .
P 2430e 0.0029t
(a) Since the exponent is negative, this is an exponential
decay model. The population is decreasing.
(b) For 2000, let t 0: P 2430 thousand people
For 2003, let t 3: P 2408.95 thousand people
Country 2000 2010
Bulgaria 7.8 7.1
Canada 31.3 34.3
China 1268.9 1347.6
United Kingdom 59.5 61.2
United States 282.3 309.2
ln 1
4
ln 1
4
ln14
3
y ae bx
1
eb3
4
ln e3b
3b
Thus, y e 0.4621x .
(c)
y ae bx
1
2 aeb0 ⇒ a 1
2
5 1
2 eb4
10 e 4b
ln 10 ln e 4b
ln 10 4b
1 ae b0 ⇒ 1 a
ln 10
b ⇒ b 0.5756
4
Thus, y 1
2 e0.5756x .
b ⇒ b 0.4621
2.3 million 2300 thousand
2300 2430e 0.0029t
2300
e0.0029t
2430
ln 2300
2430 0.0029t
t ln23002430
0.0029
18.96
The population will reach 2.3 million (according to
the model) during the later part of the year 2018.

308 Chapter 3 Exponential and Logarithmic Functions
36. —CONTINUED—
37.
(a) Bulgaria:
For 2030, use t 30.
y 7.8e million
0.009430 5.88
China:
a 7.8
7.1 7.8e b10
ln 7.1
10b ⇒ b 0.0094
7.8
a 1268.9
1347.6 1268.9e b10
ln 1347.6
10b ⇒ b 0.00602
1268.9
For 2030, use t 30.
y 1268.9e million
0.0060230 1520.06
United Kingdom:
a 59.5
61.2 59.5e b10
ln 61.2
10b ⇒ b 0.00282
59.5
For 2030, use t 30.
y 59.5e million
0.0028230 64.7
Canada:
34.3 31.3e b10
ln 34.3
10b ⇒ b 0.00915
31.3
For 2030, use t 30.
y 31.3e million
0.0091530 41.2
United States:
a 31.3
a 282.3
309.2 282.3e b10
ln 309.2
10b ⇒ b 0.0091
282.3
For 2030, use t 30.
y 282.3e million
0.009130 370.9
(b) The constant b determines the growth rates. The greater the rate of growth, the greater the value of b.
(c) The constant b determines whether the population is increasing b > 0 or decreasing b < 0.
y 4080e kt 38.
When t 3, y 10,000:
10,000 4080e k3
10,000
4080
e3k
ln 10,000
4080 3k
k ln10,0004080
3
0.2988
When y 4080e hits
0.298824 t 24:
5,309,734
y 10e kt
65 10e k14
ln 65
14k ⇒ k 0.1337
10
For 2010, t 20:
y 10e million
0.133720 $144.98

39.
41.
43.
N 100e kt 40.
300 100e 5k
3 e 5k
ln 3 ln e 5k
ln 3 5k
k
N 100e 0.2197t
200 100e 0.2197t
t
ln 3
5
0.2197
ln 2
3.15 hours
0.2197
R 1
10 12et8223
(a)
(b)
t
8223
R 1
8 14
1
1012et8223 1
814 e t8223 1012
8 14
1012
ln 814
1
1012et8223 1
1311 e t8223 1012
13 11
t 1012
ln 8223 1311 0, 30,788, 2, 18,000
(a)
m
t 8223 ln 1012
14 8 12,180 years old
t 8223 ln 1012
1311 4797 years old
Linear model:
V 6394t 30,788
—CONTINUED—
18,000 30,788
2 0
b 30,788
6394
(b)
ln 4500
2k
7697
Section 3.5 Exponential and Logarithmic Models 309
42.
N 250e kt
280 250e k10
1.12 e 10k
k
N 250eln 1.1210t
500 250eln 1.1210t
2 eln 1.1210t
ln 1.12
ln 2 10 t t
ln 1.12
10
y Ce kt
ln 1
5715k
2
ln 2
61.16
ln 1.1210
1
C Ce5715k
2
k ln12
5715
hours
The ancient charcoal has only 15% as much radioactive
carbon.
0.15C Celn 0.55715t
ln 0.15
t
ln 0.5
5715 t
5715 ln 0.15
ln 0.5
a 30,788 (c)
18,000 30,788e k2
4500
e2k
7697
k 1 4500
ln 0.268
2 7697
Exponential model: V 30,788e 0.268t
15,642 years
32,000
0 4
0
The exponential model
depreciates faster in the
first two years.

310 Chapter 3 Exponential and Logarithmic Functions
43. —CONTINUED—
44.
45.
46.
(d)
1 3
(e) The linear model gives a higher value for the car for the
first two years, then the exponential model yields a higher
$24,394 $11,606
value. If the car is less than two years old, the seller would
most likely want to use the linear model and the buyer the
V 30,788e $23,550 $13,779
exponential model. If it is more than two years old, the
opposite is true.
0.268t
t
V 6394t 30,788
0, 1150, 2, 550
(a)
(c)
m
V 300t 1150
1200
550 1150
2 0
0 4
0
300
The exponential model depreciates faster in the
first two years.
(e) The slope of the linear model means that the computer
depreciates $300 per year, then loses all value in
the third year. The exponential model depreciates
faster in the first two years but maintains value longer.
St 1001 e kt
(a)
85 100e k
85
100 ek
0.85 e k
ln 0.85 ln e k
k ln 0.85
k 0.1625
St 1001 e 0.1625t
N 301 e kt
(a)
15 1001 e k1
N 19, t 20
19 301 e 20k
30e 20k 11
e 20k 11
30
ln e 20k ln 11
30
20k ln 11
30
k 0.050
So, N 301 e 0.050 .
(b)
(d)
(b)
550 1150e k2
ln 550
2k ⇒ k 0.369
1150
Sales
(in thousands of units)
V 1150e 0.369t
t 1 3
V 300t 1100
V 1150e 0.369t
120
90
60
30
S
5 10 15 20 25 30
Time (in years)
$850 $250
$795 $380
(c) S5 1001 e 0.16255 55.625 55,625 units
(b)
N 25
25 301 e 0.050t
5
e0.050t
30
ln 5
ln e0.050t
30
ln 5
30 0.050t
t ln530
36 days
0.050
t

47.
49.
y 0.0266e x1002 450 , 70 ≤ x ≤ 116 48. (a)
(a)
0.04
70 115
0
(b) The average IQ score of an adult student is 100.
pt
1000
(a) p5
203 animals
1 9e0.16565 (b)
(c)
1 9e 0.1656t 2
1200
0
0
1000
1 9e 0.1656t
500
9e 0.1656t 1
e 0.1656t 1
9
1000
1 9e 0.1656t
t months
ln19
13
0.1656
40
The horizontal asymptotes are p 0 and p 1000.
The asymptote with the larger p-value,
p 1000,
indicates that the population size will approach 1000
as time increases.
Section 3.5 Exponential and Logarithmic Models 311
50.
0.9
4 7
0
(b) The average number of hours per week a student uses
the tutor center is 5.4.
S 500,000
1 0.6e kt
(a)
1 0.6e 4k 5
3
So, S
(b) When t 8:
500,000
1 0.6e0.0263t. k 1
4k ln
10
ln 4 9 0.0263
10
9
S
300,000 500,000
1 0.6e 4k
0.6e 4k 2
3
e 4k 10
9
500,000
287,273 units sold.
1 0.6e0.02638 51. since
(a) 7.9 log I ⇒ I 107.9 R log I0 1.
79,432,823
I
log I
I0 52. R log since I0 1.
(a) R log 80,500,000 7.91
I
log I
I0 53.
(b)
8.3 log I ⇒ I 10 8.3 199,526,231
(c) 4.2 log I ⇒ I 10 4.2 15,849
10 log I
(a)
where I0 10
I0 12 wattm2 .
10 log 1010
10 12 10 log 102 20 decibels
(c) 10 log 108
10 12 10 log 104 40 decibels
(b)
(b)
R log 48,275,000 7.68
(c) R log 251,200 5.40
10 log 105
10 12 10 log 107 70 decibels
(d) 10 log 1
10 12 10 log 1012 120 decibels

312 Chapter 3 Exponential and Logarithmic Functions
54. where I0 1012 wattm2 I 10 log I
I0 55.
57.
59.
61.
(a)
10 11 10 log 1011
10 12 10 log 101 10 decibels
(c) 10 4 10 log 104
10 12 10 log 108 80 decibels
10 log I
10
log I
I 0
10 10 10 log II 0
10 10 I
I 0
I 0
I I 0 10 10
% decrease I 0 109.3 I 0 10 8.0
I 010 9.3
100 95%
pH logH 58.
log2.3 10 5 4.64
5.8 logH 60.
5.8 logH
10 5.8 10 logH
10 5.8 H
H 1.58 10 6 mole per liter
2.9 logH 62.
2.9 logH
H 10 2.9 for the apple juice
8.0 logH
8.0 logH
H 10 8 for the drinking water
102.9 8 105.1
10
times the hydrogen ion
concentration of drinking water
T 70
63. t 10 ln
98.6 70
At 9:00 A.M. we have:
85.7 70
t 10 ln 6 hours
98.6 70
From this you can conclude that the person died at 3:00 A.M.
56.
(b)
10 2 10 log 102
10 12 10 log 1014 140 decibels
(d) 10 2 10 log 102
10 12 10 log 1010 100 decibels
10 log 10 I
10 10 I
I 0
I I 010 10
I 0
% decrease I 0 108.8 I 0 10 7.2
I 0 10 8.8
pH logH
log11.3 10 6 4.95
3.2 logH
10 3.2 H
H 6.3 10 4
pH 1 logH
pH 1 logH
10 pH1 H
10 pH1 H
10 pH 10 H
mole per liter
100 97%
The hydrogen ion concentration is increased by
a factor of 10.

64. Interest:
65.
Principal:
(a)
P 120,000, t 35, r 0.075, M 809.39
800
0
0
Pr r
u M M 1 12 12 12t
Pr r
v M 1 12 12 12t
u
v
35
(b) In the early years of the mortgage, the majority of the
monthly payment goes toward interest. The principal
and interest are nearly equal when t 26 years.
u 120,000
(a)
150,000
0
0
0.075t
1
1 12t
1 1
0.07512
66. t1 40.757 0.556s 15.817 ln s
t 2 1.2259 0.0023s 2
(a) Linear model: t3 0.2729s 6.0143
(b)
24
Exponential model: or
t4 1.53851.0296s t4 1.5385e0.02913s 25
20
0
(d) Model : t1 Model t2: Model : t3 Model : t4 t 2
t4 t1 t3 100
Section 3.5 Exponential and Logarithmic Models 313
(c) P 120,000, t 20, r 0.075, M 966.71
15.8 15.9 S 1
20 19.6 2.0
3.4 3.6 5 4.6 7 6.7 9.3 9.4 12 12.5
15.8 15.9 S4
20 21.2 2.6
3.4 3.7 5 4.9 7 6.6 9.3 8.9 S 3
15.8 15.8 20 18.5 5.6
12 11.9
3.4 2.2 5 4.9 7 7.6 9.3 10.4
12 13.1
15.8 15.9 S 2 3.4 3.3 5 4.9 7 7 9.3 9.5 12 12.5
20 19.9 1.1
The quadratic model, t2, best fits the data.
800
0
0
u
v
20
The interest is still the majority of the monthly payment
in the early years. Now the principal and interest are
nearly equal when t 10.729 11 years.
(b) From the graph, u $120,000 when t 21 years. It
would take approximately 37.6 years to pay $240,000 in
interest. Yes, it is possible to pay twice as much in interest
charges as the size of the mortgage. It is especially likely
when the interest rates are higher.
(c)
s 30 40 50 60 70 80 90
t1 3.6 4.6 6.7 9.4 12.5 15.9 19.6
t2 3.3 4.9 7.0 9.5 12.5 15.9 19.9
t3 2.2 4.9 7.6 10.4 13.1 15.8 18.5
t4 3.7 4.9 6.6 8.8 11.8 15.8 21.2
Note: Table values will vary slightly depending on the
model used for t 4 .

314 Chapter 3 Exponential and Logarithmic Functions
67. False. The domain can be the set of real numbers for a
logistic growth function.
68. False. A logistic growth function never has an x-intercept.
69. False. The graph of f x is the graph of gx shifted
70. True. Powers of e are always positive, so if a > 0, a
upward five units.
Gaussian model will always be greater than 0, and if
a < 0, a Gaussian model will always be less than 0.
71. (a) Logarithmic
73.
(b) Logistic
(c) Exponential (decay)
(d) Linear
(e) None of the above (appears to be a combination
of a linear and a quadratic)
(f) Exponential (growth)
1, 2, 0, 5 74.
(a)
(b)
−3 −2 −1 1 2 3
−1
d 0 1 2 5 2 2 1 2 3 2 10
(c) Midpoint:
(d) m
(b)
(−1, 2)
3
2
1
y
5 (0, 5)
1 0
,
2
2 5
5 2 3
3
0 1 1
−2
−2
−4
−6
−8
2 4 6 8 10 14
(14, −2)
d 14 3 2 2 3 2
(c) Midpoint:
(d) m
11 2 5 2 146
3 14
,
2
3 2
2 3
14 3 5
11
x
2 1 7
,
2 2
x
2
17
2
1
,
2
72. Answers will vary.
4, 3, 6, 1
(a)
(b)
(−6, 1)
−6 −4
2 4 6
−2
−4
−6
(4, −3)
d 6 4 2 1 3 2
(c) Midpoint:
(d) m
75. 3, 3, 14, 2 76. 10, 4, 7, 0
(a)
y
(a)
y
8
6
4
2
(3, 3)
(b)
6
4
2
y
100 16 116 229
6
4
2
−2
−2
−4
−6
(c) Midpoint:
(d) m
6 4
,
2
3 1
3 1 4
4 6 10 2
5
2
4 6 8 10
d 10 7 2 4 0 2
9 16 25 5
(7, 0)
4 0 4
10 7 3
(10, 4)
7 10
,
2
0 4
x
2 1, 1
x
2 17
2
, 2

77.
(a)
1
, 1
2
79.
81.
83.
(b)
1
−
2
d 3 1
4
(c) Midpoint:
(d) m
Line
4 , 3
4
1
1
2
1
−
2
, 0
78.
y
1
4 2
1
4 2
(( , −
1
2
1
4
3
4
2 2
(( , 0
0 1
1
8
12 34
,
2
14 0
0 14 14
1
34 12 14
Slope: m 3
y-intercept: 0, 10
1
x
4 2
2
10
8
6
4
2
−2 −2
2
6
5
8
8
, 1
8
y 10 3x y
80.
10 12
y 2x y
2 3 82.
y 2x 0 2 3
Parabola
Vertex: 0, 3
−6
2
−2
−2
3x y
2 4y 0 84.
Parabola
Vertex:
Focus:
Directrix: y 1
0,
3
1
x
0, 0
3
2 4
3 y
3x 2 4y
−4
−4 −3 −2 −1
7
6
5
4
3
2
1
1
2
2
4
3
6
4
x
x
x
Section 3.5 Exponential and Logarithmic Models 315
y
(a)
7 1
,
3 6 , 2, 1
3
(b)
−1 1 2 3
2
−
3
1
3
(( , −
d 2
7
3
(c) Midpoint:
(d) m
Line
2
1
−2
3
(( ,
7
3
32 1
2 2
9.25
23 73
,
2
13 16
y 4x 1
1
6
3 2
1 1
3 6 2
2
13 16 12 1
23 73 3 6
Slope: m 4
y-intercept: 0, 1
y 2x 2 7x 30
2x 5x 6
2x 7
4 2 289
8
Parabola
Vertex:
7 4 , 289 8
x-intercepts: 5
2 , 0, 6, 0
x 2 8y 0
Parabola
Vertex: 0, 0
Focus:
x 2 8y
0, 2
Directrix: y 2
x
5
3
2
−3 −2 −1
−1
−2
−3
y
1
,
6 12
−4 2 4 8
−5
−6
−30
−35
−4
y
2
−2
−4
−6
−8
−10
y
1
2
4
3
6
x
x
x

316 Chapter 3 Exponential and Logarithmic Functions
85.
87.
89.
91.
y
Vertical asymptote: x
Horizontal asymptote: y 0
1
3
−3
4
1 3x
−2
3
1
y
−1
−1
−2
−3
x 2 y 8 2 25
Circle
Center: 0, 8
Radius: 5
f x 2 x1 5
Horizontal asymptote:
14
12
10
8
6
4
2
y
−6 −4 −2 2 4 6 8 10
1
2
x
y 5
14
12
10
8
6
4
2
−8 −6 −4 −2
x 5 3 1 0 1 3 5
f x
f x 3 x 4
5.02 5.06 5.3 5.5 6 9 21
Horizontal asymptote: y 4
x
x 4 2 1 0 1 2
f x
3.99 3.89 3.67 3 1 5
y
2
4
6
8
x
86.
88.
90.
y
Vertical asymptote: x 2
x2
4
x 2
x 2 x 2
Slant asymptote:
10
−8 −6 −4
4
Parabola
Vertex:
P 1
4
Focus:
Directrix: y 2.75
8
6
4
2
4, 3
y
4, 3.25
y x 2
x 4 2 y 7 4
f x 2 x1 1
Horizontal asymptote:
−2
2
−4
−6
−8
−10
x 4 2 y 7 4
x 4 2 y 3
y
x
x
y 1
−2 2 4 6 8
−2
−4
−6
−8
−10
x 2 1 0 1 2
f x
3
5
4
3
2
1
−6 −5 −4 −3 −2 −1
−2
−3
−5
y
2
2
3
4
3
2
x
5
4
y
9
8
x

92.
f x 3 x 4
Horizontal asymptote: y 4
x 2 1 0 1 2
f x
3 8
9
3 2
3
93. Answers will vary.
3 1 5
Review Exercises for Chapter 3
1.
4.
7.
f x 6.1 x 2.
f 2.4 6.1 2.4 76.699
f x 1278 x5
f 1 1278 15 4.181
fx 4 x 8.
Intercept:
0, 1
Horizontal asymptote: x-axis
Increasing on: ,
Matches graph (c).
5
2
1
−5 −4 −3 −2 −1
−2
−3
−4
−5
5.
y
1 2 3 4 5
f 3 30 3 361.784
x
f x 30 x 3.
1456.529
Review Exercises for Chapter 3 317
f x 2 0.5x
f 2 0.5 0.337
f 11 70.211 f x 70.2
x 6.
f 0.8 1450.8 f x 145
3.863
x fx 4 x 9.
Intercept:
0,1
Horizontal asymptote: y 0
Decreasing on: ,
Matches graph (d).
fx 4 x
Intercept:
0, 1
Horizontal asymptote: x-axis
Decreasing on: ,
Matches graph (a).
10. fx 4
Intercept: 0, 2
Horizontal asymptote: y 1
Increasing on: ,
Matches graph (b).
x 1 11.
gx 5
Since gx fx 1, the graph
of g can be obtained by shifting
the graph of f one unit to the right.
x1
fx 5x 12. f x 4
Because gx fx 3, the
graph of g can be obtained by
shifting the graph of f three units
downward.
x , gx 4x 3
13.
15.
fx 1 2 x
gx 1
2 x2
Since gx fx 2, the graph of g can be obtained
by reflecting the graph of f about the x-axis and shifting
f two units to the left.
f x 4 x 4
Horizontal asymptote: y 4
x 1 0 1 2 3
f x
8 5 4.25 4.063 4.016
−4
−2
8
2
y
f x 2 3 x
, gx 8 2 3 x
14.
Because gx fx 8, the graph of g can be
obtained by reflecting the graph of f in the x-axis and
shifting the graph of f eight units upward.
2
4
x

318 Chapter 3 Exponential and Logarithmic Functions
16.
18.
20.
f x 4 x 3
Horizontal asymptote:
−6 −5 −4 −3 −2 −1
1
−2
−3
−4
−5
−6
−7
−8
y
1
2
3
y 3
x 2 1 0 1 2
f x
f x 2.65 x1
3.063 3.25 4 7 19
Horizontal asymptote:
5
4
3
2
1
y
−3 −2 −1
−1
1
2
3
x
y 0
x 3 1 0 1 3
f x
f x 2 x6 5
0.020 0.142 0.377 1 7.023
Horizontal asymptote:
6
4
2
y
−2 2 4 6 10
−2
−4
−6
x
y 5
x 0 5 6 7 8 9
f x
4.984 4.5 4 3 1 3
x
17.
19.
21.
f x 2.65 x1
Horizontal asymptote:
−6 −3 3 6 9
−3
−6
−9
−12
−15
y
y 0
x 2 1 0 1 2
f x
0.377
f x 5 x2 4
1
Horizontal asymptote:
−4
−2
8
6
2
y
2
4
2.65
x
x
y 4
x 1 0 1 2 3
f x
4.008
4.04
4.2
f x 1 2 x
3 2x 3
Horizontal asymptote:
−4
−2
8
6
2
y
2
y 3
x 2
1 0 1 2
f x
3.25
3.5
4 5 7
4
x
5 9
7.023
18.61

22.
23.
f x 1 8 x2
5
Horizontal asymptote: y 5
x 3 2 1 0 2
f x
3 x2 1
9
3 x2 3 2
x 2 2
x 4
3 4 4.875 4.984 5
24.
1 3 x2
1 3 x2
1 3 x2
1 3 4
x 2 4
x 2
−4
81 25.
3 4
2
−2
−4
−6
y
2
4
5x 7 15
5x 22
x 22
5
Review Exercises for Chapter 3 319
x
e 5x7 e 15 26.
e 82x e 3
8 2x 3
2x 11
x 11
2
27. e 8 2980.958 28. e 58 1.868 29. e 1.7 0.183 30. e 0.278 1.320
31.
33.
hx e x2
x 2 1 0 1 2
hx
2.72 1.65 1 0.61 0.37
7
6
5
4
3
2
−4 −3 −2 −1 1 2 3 4
y
f x e x2
x 3 2 1 0 1
f x
0.37
−6 −5 −4 −3 −2 −1 1 2
7
6
2
1
x
1 2.72 7.39 20.09
y
x
32.
34.
hx 2 e x2
x 2 1 0 1 2
y 0.72 0.35 1 1.39 1.63
3
−4 −3 −1 1 2 3 4
−2
−3
−4
−5
st 4e 2t , t > 0
5
4
3
2
1
y
1
2
y
t 1 2 3 4
y 0.07 0.54 1.47 2.05 2.43
1 2 3 4 5
x
t

320 Chapter 3 Exponential and Logarithmic Functions
0.065
35. A 35001
n 10n
or A 3500e0.06510 n 1 2 4 12 365
Continuous
Compounding
A $6569.98 $6635.43 $6669.46 $6692.64 $6704.00 $6704.39
0.05
36. A 20001
n 30n
or A 2000e0.0530 n 1 2 4 12 365 Continuous
A $8643.88 $8799.58 $8880.43 $8935.49 $8962.46 $8963.38
37.
(a) F (b) F2 0.487 (c) F5 0.811
1
Ft) 1 e
2 0.154
t3
38.
40.
41.
44.
Vt 14,0003 4 t
(a)
15,000
0
0
10
(b) V2 14,000
(c) According to the model, the car depreciates most
rapidly at the beginning. Yes, this is realistic.
3
4 2 $7875
Q 1001 2 t14.4
39. (a)
A 50,000e 0.087535 $1,069,047.14
(b) The doubling time is
ln 2
7.9 years.
0.0875
(a) For grams (b) For t 10: Q 100 grams
1
1
t 0: Q 100
(c)
Mass of 241 Pu (in grams)
100
80
60
40
20
Q
log 4 64 3
t
20 40 60 80 100
Time (in years)
2 014.4 100
4 3 64 42.
e 0 1 45.
ln 1 0
log25 125 3
25
2
32 125 43.
f 1000 log 1000
log 10 3 3
2 1014.4 61.79
e 0.8 2.2255 . . .
ln 2.2255 . . . 0.8
f x log x 46. log 9 3 log 9 9 12 1
2

47.
50.
g1 8 log21 8 log2 23 gx log2 x
3
3x 10 5
3x 15
x 5
48.
log 83x 10 log 8 5 51.
53. gx log7 x ⇒ x 7
Domain: 0,
y
55.
x-intercept:
1, 0
Vertical asymptote: x 0
x 1 7 49
gx
1
7
1
f x log
Domain: 0,
x
3
x-intercept:
3, 0
Vertical asymptote:
x 0
0 1 2
x 0.03 0.3 3 30
f x
2
4
3
2
1
y
−2 −1
−1
−2
⇒ x
3 10y ⇒ x 310 y
1
57. f x 4 logx 5
Domain: 5,
x-intercept:
9995, 0
0 1
3
2
1
−1
−1
−2
−3
Since 4 logx 5 0 ⇒ logx 5 4
Vertical asymptote: x 5
y
1
2
f 1 4 log4 1
fx log4 x 49.
4 1
2
lnx 9 ln 4 52.
3
x 5 10 4
x 10 4 5 9995.
3
x 9 4
4
4
x 5
5
x
x
54. gx log5 x ⇒ 5
Domain: 0,
y x
log 5 x 0
x 5 0
x 1
x-intercept:
Vertical asymptote: x 0
Review Exercises for Chapter 3 321
1, 0
log 4x 7 log 4 14
x 7 14
x 7
ln2x 1 ln11
2x 1 11
x 1 5 25
gx
1
25
1
5
56. f x 6 log x
Domain: 0,
6 log x 0
x-intercept:
2 1 0 1 2
log x 6
x 10 6
x 0.000001
0.000001, 0
Vertical asymptote: x 0
2x 12
x 6
3
2
1
−1 1 2 3 4 5
−1
−2
−3
10
8
6
4
2
−2
−2
x 1 2 4 6 8 10
f x
x 4
3 2 1 0 1
f x
4 3.70 3.52 3.40 3.30 3.22
6 6.3 6.6 6.8 6.9 7
−6
y
y
2
−4 −3 −2 −1
7
6
5
4
3
2
1
y
4
1
6
2
8
x
10
x
x

322 Chapter 3 Exponential and Logarithmic Functions
58. f x logx 3 1
Domain: 3,
logx 3 1 0
logx 3 1
x-intercept:
x 3 10 1
x 3.1
3.1, 0
Vertical asymptote: x 3
x 4 5 6 7 8
f x
1 1.3 1.5 1.6 1.7
59. ln 22.6 3.118 60. ln 0.98 0.020 61. ln e 12 12
62. ln e7 7 63. ln7 5 2.034 64. ln 3
8 1.530
65. fx ln x 3
Domain: 0,
x-intercept:
ln x 3 0
ln x 3
x e 3
e 3 , 0
Vertical asymptote: x 0
x 1 2 3
f x
1
2
3 3.69 4.10 2.31 1.61
67. hx lnx
Domain: , 0 0,
x-intercepts: ±1, 0
2 2 lnx 69.
Vertical asymptote: x 0
x
±0.5
±1
±2
6
5
4
3
2
1
y
−1 1 2 3 4 5
1
4
−4 −3 −2 −1
±3
±4
4
3
2
1
−3
−4
y 1.39 0 1.39 2.20 2.77
h 116 loga 40 176 70.
h55 116 log55 40 176
53.4 inches
y
1
2
3
4
x
s 25
27.16 miles
x
13 ln1012
ln 3
66. fx lnx 3
Domain: 3,
lnx 3 0
x 3 e 0
x-intercept:
x 4
4, 0
−1
5
4
3
2
1
−2
−3
−4
−5
Vertical asymptote: x 3
x 3.5 4 4.5 5 5.5
y 0.69 0 0.41 0.69 0.92
68. fx
Domain: 0,
1
4 ln x
1
4 ln x 0
ln x 0
x e 0
x 1
x-intercept:
Vertical asymptote: x 0
1
2
1, 0
3
2
y
1
2
3
2
1
−1
−2
−3
y
4
4
2
−2
−4
y
5
6
7
8
9
2 4 6 8
1 2 3 4 5 6
x 1 2 3
y 0.17 0 0.10 0.17 0.23 0.27
71.
log 4 9
log 4 9
5
2
log 9
1.585
log 4
ln 9
1.585
ln 4
x
x
x

72.
75.
77.
80.
83.
85.
log 12 200
log 12 200
1.255
log 200
log 12
ln 200
ln 12
2.132 73.
2.132
log 2 2 log 3
log 12 5
log 12 5
log 5
2.322 74.
log12
ln 5
2.322
ln12
log 18 log2 3 2 76.
ln 20 ln2 2 5
2 ln 2 ln 5 2.996
log 10 7x 4 log 7 log x 4 81.
log 7 4 log x
2 ln x 2 ln y ln z
3.585
Review Exercises for Chapter 3 323
log 3 0.28
log 3 0.28
log 0.28
log 3
ln 0.28
ln 3
log 2 1
12 log 2 1 log 2 12 0 log2 2 3
2 log 2 2 2 log 2 3 2
1.159
1.159
log 3
log 2
78. ln 3e4 ln 3 ln e4 79. log5 5x2 log5 5 log5 x2 ln 3 4
2.90
ln x 2 y 2 z ln x 2 ln y 2 ln z 84.
lnx 3 ln x ln y
lnx 3 ln x ln y
log 3 6
3
x log 3 6 log 3 3 x 82.
log 33 2 log 3 x 13
log 3 3 log 3 2 1
3 log 3 x
1 log 3 2 1
3 log 3 x
x 3
ln xy lnx 3 ln xy 86.
ln 3xy 2 ln 3 ln x ln y 2
ln 3 ln x 2 ln y
y 1
ln 4 2 y 1
2 ln 4
1 2 log 5 x
log 7 x
4 log 7 x log 7 4
2 lny 1 2 ln 4
log 7 x 12 log 7 4
1
2 log 7 x log 7 4
2 lny 1 ln 16, y > 1
87. log2 5 log2 x log2 5x 88. log6 y 2 log6 z log6 y log6 z2 y
89. ln x 1
4 ln y ln x ln 4 y ln x
4
91.
1
3 log 8x 4 7 log 8 y log 8 3 x 4 log8 y 7 92.
log 8y 7 3 x 4
log 6 y
z 2
90. 3 ln x 2 lnx 1 ln x3 lnx 12 ln x 3 x 1 2
2 log x 5 logx 6 log x 2 logx 6 5
x
log
2
x 65 1
log
x2x 65

324 Chapter 3 Exponential and Logarithmic Functions
93.
94. 5 lnx 2 lnx 2 3 ln x lnx 25 lnx 2 ln x3 95. t 50 log
(a) Domain: 0 ≤ h < 18,000
18,000
18,000 h
(b)
100
0 20,000
0
Vertical asymptote: h 18,000
96. Using a calculator gives
100.
104.
108.
1
ln2x 1 2 lnx 1 ln2x 1 lnx 12
2
s 84.66 11 ln t.
e x 6 101.
ln e x ln 6
x ln 6 1.792
ln x 3 105.
x e 3 0.0498
e 3x2 40
ln e 3x2 ln 40
3x 2 ln 40
x
ln 40 2
3
2x 1
ln
x 12 lnx 2 5 lnx 2 ln x 3
lnx 2 5 ln x 3 x 2
x 25
ln
x3x 2
97. 8 x 512 98.
8 x 8 3
x 3
log 4 x 2 102.
x 4 2 16
ln e x ln12
14e 3x2 560 109.
0.563
e x 12 106.
x ln 12 2.485
(c) As the plane approaches its absolute ceiling, it climbs at
a slower rate, so the time required increases.
18,000
(d) 50 log
5.46 minutes
18,000 4000
6 x 1
216
6 x 6 3
x 3
6 log 6 x 6 1
x 1
6
99.
e x 3
x ln 3
log6 x 1 103. ln x 4
e 3x 25 107.
ln e 3x ln 25
3x ln 25
x
ln 25
3
1.073
2 x 13 35 110.
2 x 22
x log 2 22
log 22
log 2
x 4.459
or ln 22
ln 2
6 x 28 8
6 x 20
x
x e 4
e 4x e x2 3
4x x 2 3
log 6 6 x log 6 20
0 x 2 4x 3
0 x 1x 3
x 1 or x 3
x log 6 20
ln 20
ln 6
1.672

111.
113.
115.
117.
45 x 68 112.
5 x 17
ln 5 x ln 17
x ln 5 ln 17
x
e 2x 7e x 10 0 114.
e x 2e x 5 0
e x 2 or e x 5
ln e x ln 2 ln e x ln 5
x ln 2 0.693 x ln 5 1.609
2 10
0.6x 3x 0 116.
Graph y1 2 .
0.6x 3x
The x-intercepts are at
x 0.392 and at x 7.480.
Graph y1 25e and
y2 12.
0.3x
The graphs intersect at
x 2.447.
−10
25e 16
0.3x 12 118.
e ln 3x e 8.2
3x e 8.2
x e8.2
3
1213.650
−6
−2
−10
10
18
5x e 7.2
x e7.2
5
267.886
e 2x 6e x 8 0
e x 2e x 4 0
e x 2 or e x 4
x ln 2 x ln 4
Review Exercises for Chapter 3 325
x 0.693 x 1.386
4 0.2x x 0
Graph y1 4 .
0.2x x
The x-intercepts are at
x 7.038 and at
x 1.527.
4e 1.2x 9
Graph y1 4e and y2 9.
1.2x
The graphs intersect at
x 0.676.
119. ln 3x 8.2 120. ln 5x 7.2 121. 2 ln 4x 15
ln 3x 15
4
ln 17
ln 5
3x e 154
x e154
3
1.760
14.174
ln x
2
3
e lnx3 e 2
x
e2
3
x 3e 2 22.167
212 x 190
12 x 95
ln 12 x ln 95
x ln 12 ln 95
122. 4 ln 3x 15 123. ln x ln 3 2 124.
x
ln 95
1.833
ln 12
−12 6
−6
ln 4x 15
2
e ln 4x e 7.5
4x e 7.5
12
−2
9
−3
x 1
4 e7.5 452.011
lnx 8 3
1
lnx 8 3
2
lnx 8 6
x 8 e 6
x e 6 8 395.429
6

326 Chapter 3 Exponential and Logarithmic Functions
125.
127.
129.
log 8x 1 log 8x 2 log 8x 2 128.
log 8x 1 log 8 x 2
x 2
x 1
x 1x 2 x 2
x 2 x 2 x 2
x 2 0
x 0
x 2
x 2
Since x 0 is not in the domain of log8x 1 or of
log8x 2, it is an extraneous solution. The equation
has no solution.
log1 x 1 130.
1 x 10 1
1 1
10 x
x 0.900
log 6x 2 log 6 x log 6x 5
log 6 x 2
x log 6x 5
x 2
x 5
x
x 2 x 2 5x
0 x 2 4x 2
x 2 ± 6, Quadratic Formula
Only x 2 6 0.449 is a valid solution.
logx 4 2
x 4 10 2
x 100 4
x 104
131. 2 lnx 3 3x 8 132.
Graph y1 2 lnx 3 3x and y2 8.
Graph y1 6 logx2 6 logx
1 x.
2 1 x 0
133.
lnx 1 2 126. ln x ln 5 4
1
lnx 1 2
2
10
−9 9
−2
(1.64, 8)
The graphs intersect at approximately 1.643, 8.
The solution of the equation is x 1.643.
Graph y1 4 lnx 5 x and y2 10.
−6
lnx 1 4
e lnx1 e 4
x 1 e 4
4 lnx 5 x 10
11
−1
x e 4 1 53.598
12
The graphs do not intersect. The equation has no solution.
ln x
4
5
12
−8 16
−4
x
e4
5
x 5e 4 272.991
The x-intercepts are at x 0, x 0.416, and x 13.627.

134.
136.
138.
x 2 logx 4 0 135.
Let y1 x 2 logx 4.
12
−8 16
−4
The x-intercepts are at x 3.990 and x 1.477.
283 93 logd 65
218 93 logd
logd 218
93
d 10 21893 220.8 miles
37550 7550e 0.0725t
3 e 0.0725t
ln 3 ln e 0.0725t
ln 3 0.0725t
S 93 logd 65 137. y 3e2x3 y 4e 2x3 139.
Exponential growth model
Matches graph (b).
t
Review Exercises for Chapter 3 327
ln 3
15.2 years
0.0725
Exponential decay model
Matches graph (e).
y lnx 3
Logarithmic model
Vertical asymptote: x 3
Graph includes 2, 0
Matches graph (f).
140. y 7 logx 3 141. y 2e
Logarithmic model
Gaussian model
Vertical asymptote:
Matches graph (d).
x 3
Matches graph (a).
x423 142.
6
y
1 2e
Logistics growth model
Matches graph (c).
2x
143.
y ae bx 144.
2 ae b0 ⇒ a 2
3 2e b4
1.5 e 4b
ln 1.5 4b ⇒ b 0.1014
Thus, y 2e 0.1014x .
y ae bx
1
2 aeb0 ⇒ a 1
2
5 1
2 eb5
10 e 5b
ln 10 5b
ln 10
b
5
b 0.4605
y 1
2 e0.4605x

328 Chapter 3 Exponential and Logarithmic Functions
145.
147. (a)
P 3499e 0.0135t 146.
4.5 million 4500 thousand
4500 3499e 0.0135t
4500
e0.0135t
3499
ln 4500
0.0135t
3499
t ln45003499
0.0135
18.6 years
According to this model, the population of South
Carolina will reach 4.5 million during the year 2008.
(b)
13.8629%
A 10,000e 0.138629
$11,486.98
157
150. N
1 5.4e
(a) When N 50:
0.12t
50
1 5.4e 0.12t 157
50
5.4e 0.12t 107
50
e 0.12t 107
270
0.12t ln 107
270
157
1 5.4e 0.12t
t ln107270
0.12
7.7 weeks
k ln710
3
The population one year ago:
N4 2000e 0.118894
1243 bats
y Ce kt
1
C Ce250,000k
2
ln 1
ln e250,000k
2
ln 1
250,000k
2
k ln12
250,000
When t 5000, we have
y Ce ln12250,0005000 0.986C 98.6%C.
After 5000 years, approximately 98.6% of the
radioactive uranium II will remain.
2 e
ln 2 5r
ln 2
r
5
r 0.138629
5r
20,000 10,000er5 148. and so
and:
3k ln 7
10
7
1400 2000e
e3k
10 3k
N 2000ekt N0 2000 N3 1400 149.
(a) Graph y1 0.0499e
0.05
x712 y 0.0499e
40 ≤ x ≤ 100
128 .
x712128 ,
0.11889
(b) When N 75:
75
1 5.4e 0.12t 157
75
5.4e 0.12t 82
75
e 0.12t 82
405
0.12t ln 82
405
157
1 5.4e 0.12t
t ln82405
0.12
40 100
0
(b) The average test score is 71.
13.3 weeks

151.
10 log
I
125 10 log1016
I
12.5 log1016
10 12.5 I
10 16
I
1016 I 10 3.5 wattcm 2
152.
R log I since I 0 1.
(a)
(b)
(c)
log I 8.4
log I 6.85
log I 9.1
Problem Solving for Chapter 3 329
I 10 8.4 251,188,643
I 10 6.85 7,079,458
I 10 9.1 1,258,925,412
153. True. By the inverse properties, log b b 2x 2x. 154. False. ln x ln y lnxy lnx y
155. Since graphs (b) and (d) represent exponential decay, b and d are negative.
Since graph (a) and (c) represent exponential growth, a and c are positive.
Problem Solving for Chapter 3
1.
y a y
x 2.
y 1 0.5 x
y 2 1.2 x
y 3 2.0 x
y 4 x
−4 −3 −2 −1 1 2
The curves and cross the line
From checking the graphs it appears that will cross
y a for 0 ≤ a ≤ 1.44.
x
y 1.2 y x.
y x
x
y 0.5x 7
6
5
4
3
2
y 3
y 4
y 1
y 2
3 4
x
y 1 e x
y 2 x 2
y 3 x 3
y5 x y4 x
0 6
0
The function that increases at the fastest rate for “large”
values of x is (Note: One of the intersection
points of and is approximately
and past this point e This is not shown on the
graph above.)
x > x3 y x 4.536, 93
.
3
y ex y1 ex .
3. The exponential function, increases at a faster rate
than the polynomial function y xn y e
.
x , 4. It usually implies rapid growth.
5. (a) fu v auv 6.
(b) f2x a2x 7. (a)
a u a v
fu fv
a x 2
fx 2
6
y = e x y 1
−6 6
−2
(b)
y 2
y = e x
−2
f x2 gx2 ex ex 2
e2x 2 e 2x
4
4
4
1
6 (c)
−6 6
2
y = e x
24
y 3
y 1
y 5
ex ex 2 2
6
−6 6
y3 −2
y 2
y 4
e2x 2 e2x
4

330 Chapter 3 Exponential and Logarithmic Functions
8.
10.
y4 1 x x2 x3 x4
1! 2! 3! 4!
y 4
6
−6 6
−2
y = e x
As more terms are added, the polynomial approaches
ex 1 x
e
x2 x3 x4 x5
. . .
1! 2! 3! 4! 5! x .
x ay 1
a y 1
xa y 1 a y 1
xa y a y x 1
a y x 1 x 1
a y
x 1
x 1
y log a x 1
x 1
x 1
lnx 1
f
ln a
1x 9.
e 2y xe y 10
e y x ± x2 4
2
fx ax 1
ax , a > 0, a 1 11. Answer (c).
1
12. (a) The steeper curve represents the investment earning compound interest,
because compound interest earns more than simple interest. With simple
interest there is no compounding so the growth is linear.
(b) Compound interest formula:
Simple interest formula:
A 5001 0.07
1 1t 5001.07 t
A Prt P 5000.07 t 500
(c) One should choose compound interest since the earnings would be higher.
2 tk 1
1
13. y1 c1 and
1
c12 tk1 1
c22 tk2 c1
c 2
1
t
1
y2 c22 tk2 2 tk 2 tk 1
ln c1 c2 t
k2 t 1
k ln
1 2
ln c1 ln c2 t 1
k2 1 1
k ln
1 2
ln c 1 ln c 2
1k 2 1k 1 ln12
fx e x e x
y e x e x
x e y e y
x e2y 1
e y
xe y e 2y 1
Quadratic Formula
Choosing the positive quantity for we have
Thus, f 1x ln x x2 4
2 .
y ln x x2 4
2 .
y 61 e
The graph passes through 0, 0 and neither (a) nor (b) pass
through the origin. Also, the graph has y-axis symmetry and
a horizontal asymptote at y 6.
x22 Growth of investment
(in dollars)
4000
3000
2000
1000
A
e y
Compounded Interest
Simple Interest
5 10 15 20 25 30
Time (in years)
14. B B0a through 0, 500 and 2, 200
kt
B 0 500
200 500a k2
2
a2k
5
2
loga 2k 5
1
2 loga 2
k 5
−4 −3 −2 −1 1 2
B 500a 12 loga25t 500a loga 25 t2 500 2
5 t2
4
3
2
1
−4
t
y
3 4
x

15. (a)
17.
18.
(b)
(c)
y 252.6061.0310 t
y 400.88t 2 1464.6t 291,782
2,900,000
y 2
y 1
0 85
200,000
(d) Both models appear to be “good fits” for the data, but
neither would be reliable to predict the population of
the United States in 2010. The exponential model
approaches infinity rapidly.
ln x 2 ln x 2
ln x 2 2 ln x 0
ln xln x 2 0
ln x 0 or ln x 2
x 1 or x e 2
y ln x
y 1 x 1
y 2 x 1 1
2x 1 2
y 3 x 1 1
2x 1 2 1
3x 1 3
(a) 4
(b) 4
(c)
−4
y 1
y = ln x
−3 9
y = ln x
−3 9
y2 19.
The pattern implies that
ln x x 1 1
2x 12 1
3x 13 1
4x 14 y
. . . .
4 x 1 1
2x 12 1
3x 13 1
4x 14 20.
ln y lnab x
ln y ln a ln b x
ln y ln a x ln b
ln y ln bx ln a
Slope: m ln b
y-intercept: 0, ln a
−4
Problem Solving for Chapter 3 331
16. Let and Then and
x abn x a
.
m
loga x m logab x n.
am a
b n
a mn a
b
a mn1 1
b
loga 1 m
1
b n
1 loga 1 m
b n
1 log a 1
b log a x
log ab x
4
−4
y 3
y = ln x
−3 9
4
−3 9
−4
y = ln x
y abx y ax 21. y 80.4 11 ln x
b
ln y lnax b
ln y ln a ln x b
ln y ln a b ln x
ln y b ln x ln a
Slope: m b
y-intercept: 0, ln a
30
y 4
100 1500
0
y300 80.4 11 ln 300 17.7 ft 3 min

332 Chapter 3 Exponential and Logarithmic Functions
22. (a) cubic feet per minute
(b)
ln x 65.4
450
15
30
(c) Total air space required: 38230 11,460 cubic feet
15 80.4 11 ln x
Let x floor space in square feet and h 30 feet.
11 ln x 65.4
V xh
11
11,460 x30
x 382
23. (a)
x e 65.411
x 382 cubic feet of air space per child.
(b) The data could best be modeled by a logarithmic
model.
(c) The shape of the curve looks much more logarithmic
than linear or exponential.
(d)
25. (a)
9 24. (a)
0 9
0
y 2.1518 2.7044 ln x
9
0 9
0
(e) The model is a good fit to the actual data.
(b) The data could best be modeled by a linear model.
(c) The shape of the curve looks much more linear than
exponential or logarithmic.
(d)
9 26. (a)
0 9
0
y 0.7884x 8.2566
9
0 9
0
(e) The model is a good fit to the actual data.
If the ceiling height is 30 feet, the minimum number of
square feet of floor space required is 382 square feet.
(b) The data could best be modeled by an exponential
model.
(c) The data scatter plot looks exponential.
(d)
36
0 9
0
y 3.1141.341 x
36
0 9
0
(e) The model graph hits every point of the scatter plot.
(b) The data could best be modeled by a logarithmic
model.
(c) The data scatter plot looks logarithmic.
(d)
10
0 9
0
y 5.099 1.92 lnx
10
0 9
0
(e) The model graph hits every point of the scatter plot.

Chapter 3 Practice Test
1. Solve for x: x 35 8.
2. Solve for x: 3 x1 1
81 .
3. Graph fx 2 x .
4. Graph gx e x 1.
5. If $5000 is invested at 9% interest, find the amount after three years if the interest is compounded
(a) monthly. (b) quarterly. (c) continuously.
6. Write the equation in logarithmic form: 7 2 1
49 .
7. Solve for x: x 4 log 2 1
64 .
8. Given evaluate logb 4 logb 2 0.3562 and logb 5 0.8271, 825.
9. Write 5 ln x as a single logarithm.
1
ln y 6 ln z
2
10. Using your calculator and the change of base formula, evaluate log 9 28.
11. Use your calculator to solve for N: log 10 N 0.6646
12. Graph y log 4 x.
13. Determine the domain of fx log 3x 2 9.
14. Graph y lnx 2.
15. True or false:
16. Solve for x: 5 x 41
ln x
lnx y
ln y
17. Solve for x: x x 2 log 5 1
25
18. Solve for x: log 2 x log 2x 3 2
19. Solve for x: ex e x
3
4
20. Six thousand dollars is deposited into a fund at an annual interest rate of 13%. Find the time required
for the investment to double if the interest is compounded continuously.
Practice Test for Chapter 3 333