Section 12.5 Exponential Growth and Decay; Modeling Data
29. First find the decay equation.
22
22
0.5
0.5
ln 0.5 ln
kt
k
k
e
e
e
30. First find the decay equation.
12
12
0.5
0.5
ln 0.5 ln
kt
k
k
e
e
e
31. First find the decay equation.
36
36
0.5
0.5
ln 0.5 ln
kt
k
k
e
e
e
32.
0
5
5
1000 1400
1000
1400
kt
k
k
AAe
e
e
Chapter 12 Exponential and Logarithmic Functions
33.
00
2
2
kt
kt
A
Ae
e
34.
0
00
3
3
kt
kt
kt
A
Ae
A
Ae
e
35.
0.01
4.3
t
Ae
a. 0.01k, so New Zealand’s growth rate is 1%.
b.
0.01
4.3
t
Ae
36.
0.012
112.5
t
Ae
a. 0.012k, so Mexico’s growth rate is 1.2%.
b.
0.012
112.5
t
Ae
37. a.
38. a.
39. a.
Section 12.5 Exponential Growth and Decay; Modeling Data
41. a.
b.
A logarithmic function appears to be
the best choice for modeling the data.
42. a.
43.
ln 4.6
1.526
100 4.6
100
100
x
x
x
y
ye
ye
46.
ln0.6
0.511
4.5 0.6
4.5
4.5
x
x
x
y
ye
ye
b.
ln1.011
0.0109
201.2 1.011
201.2
201.2
x
x
x
y
ye
ye
57. The linear model is 2.657 197.923yx . Since
0.997r is close to 1, the model fits the data very
well.
58. The power regression model is
0.100
195.05yx.
Since r = 0.896, the model fits the data fairly well.
201.2
335
ln ln 1.011
201.2
335
ln ln 1.011
x
x
2.657
x
1969 52 2021
According to the exponential model, the U.S.
population will reach 335 million around the year
Chapter 12 Exponential and Logarithmic Functions
60. a. Exponential Regression:
3.46(1.02) ; 0.994
x
yr
b.
3.46(1.02)
x
y
61. Models and predictions will vary. Sample models
are provided
Exercise 37:
1.402 1.078
x
y
62. does not make sense; Explanations will vary.
Sample explanation: Since the car’s value is
decreasing (depreciating), the growth rate is
negative.
63. does not make sense; Explanations will vary.
64. makes sense
70. Use
0
210T
,
70C
,
30t
, and
140T
to
determine the constant k:
70 140
140 140
30
e
Thus, the model for these conditions is
0.0231
70 140
t
Te
.
22
22
22
2732115
92115
273 3
3
xx x x
xxx
xx xx
x
3x
21x
3x
21x
5
3
x
xx
72.
33
230xx
1
3
Chapter 12 Review Exercises
Substitute
1
3
for .xt
73.
62 250 398
74.
2
4 xx
++
22
44(2)xx x
++=+
76. Graph of Circle:
Chapter 12 Review Exercises
1.
4
x
fx
x
fx
−2
2
2
11
416
4
2.
4
x
fx
1
1
1
11
44
4
3.
4
x
fx
x
fx
0
00
441
1
1
1
11
44
4.
43
x
fx
x
fx
2
22
434316313
1
11
4343431
0
00
4343132
Chapter 12 Exponential and Logarithmic Functions
5.
1
2 and 2
xx
fx gx
x
fx
gx
−2
1
1
The graph of g is the graph of f shifted 1 unit to the
right.
6.
1
2 and 2
x
x
fx gx
x
fx
gx
−2
1
4
7.
3 and 3 1
xx
fx gx
x
fx
gx
−2
1
8
The graph of g is the graph of f shifted down 1 unit.
8.
3 and 3
xx
fx gx
x
fx
gx
−2
1
9
1
9
−1
1
1
The graph of g is the graph of f reflected across the
Chapter 12 Review Exercises
9. 5.5% Compounded Semiannually:
25
0.055
5000 1 2
A
5.5% compounded semiannually yields the greater
return.
10. 7.0% Compounded Monthly:
12 10
0.07
14000 1 12
A
11. a. The coffee was
200
F
when it was first taken
out of the microwave.
12.
49
1
2
1log 7
2
49 7
14.
3
log 81
381
y
y
18.
3
44
log 64 log 4 3
because
log
x
b
bx
.
19.
2
555
2
11
log log log 5 2
25 5
because
log
x
b
bx
.
20.
3
log 9
22.
17
log 17 1
because
1
17 17
.
8
x
log
x
b
bx
.
26.
2
2
1
ln ln 2e
e
because
log
x
b
bx
.
Chapter 12 Exponential and Logarithmic Functions
28. Recall that
log 1
b
b
and
log 1 0
b
for all
0b
,
1b
. Therefore,
38 3
log log 8 log 1 0
.
29.
2
2; log
x
fx gx x
Domain of g: {| 0}xx or
0,
Range of g: { | is a real number}yy or
,
.
30.
1
3
1; log
3
x
fx gx x
31.
8
log 5fx x
50
x
32.
log 3fx x
30
3
3
x
x
x
34. Since ln ,
x
ex
6
ln 6 .
x
ex
35. Since
ln
,
x
ex
ln
.
x
ex
36. Since
log
10 ,
x
x2
log 4 2
10 4 .
x
x
37.
0
0
log
1000
I
RI
I
given was 76.
b.
27618log21
76 18log 3 67.4
f
Chapter 12 Review Exercises
87618log81
f
The average scores were as follows:
2 months 67.4
c.
39. 112
ln
0.06 12 5
112
ln 9.0
0.06 7
t
It will take approximately 9 weeks for the man to
run 5 miles per hour.
40.
33
666 6
log 36 log 36 log 2 3logxxx
1
ln 1 ln
333
xx
44.
log 7 log 3 log 7 3 log 21
bb b b
46.
34 34
3ln 4 ln ln ln lnxyxy xy
47.
1
2
1ln ln ln ln
2
xyx y
x
49.
4
log 0.863 0.1063
log 4
50. true
51. false; Changes to make the statement true will vary.
A sample change is:
9
log( 9) log( 1) log 1
x
xx x
.
52. false; Changes to make the statement true will vary.
Chapter 12 Exponential and Logarithmic Functions
55.
32
125 25
55
x
x
56.
23
1
927
33
x
x
57. 812,143
ln8 ln12,143
x
x
58.
5
5
9 1269
1269
x
x
e
e
59.
0.045
0.045
30 90
90
x
x
e
e
60.
5
3
log 3
5
1
x
x
The solution set is
100 .
62.
4
log 3 5 3
x
63.
1
ln 1
1
x
xe
Chapter 12 Review Exercises
65.
22
2
2
2
log 3 log 3 4
log 3 3 4
log 9 4
xx
xx
x
66.
33
log 1 log 2 2xx
3
1
log 2
2
x
x
We disregard
19
8
because it would result in
taking the logarithm of a negative number in the
original equation. There is no solution. The
solution set is
or .
67.
44
log (3 5) log 3
x
68. ln( 4) ln( 1) lnxxx
4
ln ln
1
xx
x
We disregard –2 because it would result in taking
the logarithm of a negative number in the original
equation. The solution set is
2.
70.
0.21
14.7 x
Px e
0.21
0.21
0.21
4.6 14.7
4.6
14.7
4.6
ln ln
14.7
x
x
x
e
e
e