11.1 SOLUTIONS
115
CHAPTER ELEVEN
Solutions for Section 11.1
EXERCISES
1. We use the fact that if y = log x then 10y = x. This gives 10−2 = 0.01.
5. We use the fact that if y = log x then 10y = x. This gives 103.699 = 5000.
9. We use the fact that if 10y = x then y = log x. This gives log 100,000 = 5.
13. We use the fact that 100 = 102 , so 1002.301 = 102·2.301 = 104.602 . Then we use the fact that if 10y = x then y = log x.
This gives log 39,994 = 4.602.
17. Taking logs, we have x = log 10,000 = 4. However, if we notice that 10,000 is the 4th power of 10, we don’t need to
use logs.
21. Taking logs we have x = log 17.717 = 1.248.
25. We have log 0.001 = −3 because 10−3 = 0.001.
29. Since 10 to any real power is positive, −100 cannot be written as a power of 10. So log(−100) is undefined.
33. Since log 10N = N , we have log 107 = 7.
PROBLEMS
37. (a) We have log 2 = 0.301, log 20 = 1.301, log 200 = 2.301, and log 2000 = 3.301. It seems that log(2 · 10n ) =
n + 0.301.
(b) We have log 20,000 = log(2 · 104 ) = 4.301 and log 0.2 = log(2 · 10−1 ) = −1 + 0.301 = −0.699.
(c) Using several properties of logarithms, log(2 · 10n ) = log 2 + log 10n = log 2 + n log 10 = log 2 + n = n + 0.301,
which is what we guessed in part (a).
41. Since
1 · 104 <
0.99 · 105
5
< 1 · 105
4 < log(0.99 · 10 ) < 5.
45. Since 101.42 = 26.3 and 101.44 = 27.5, we know log 27 is between 1.42 and 1.44.
49. From Table 11.1 we see that an approximate solution is x = −1.5. Using logs gives
10x = 0.03
x = log 0.03 = −1.5229.
Table 11.1
x
−1.6
−1.5
−1.4
−1.3
10x
0.025
0.032
0.040
0.050
53. Yes, since log(10b) = log 10 + log b = 1 + log b.
57. We have log(2A/B) = log(2A) − log B = log 2 + log A − log B.
61. Not possible.
116
Chapter Eleven /SOLUTIONS
65. We have log
x
z
= log x − log z = u − w.
69. We can not use log properties to rewrite this, because no log property allows us to break apart terms combined by addition.
73. This roughly resembles the graph of y = log x. However, it contains the approximate point (x, y) = (3, 1), whereas the
graph of y = log x contains the point (3, log 3) = (3, 0.477). So this graph could be a vertically stretched version of the
log graph, perhaps given by the formula y = 2 log x.
Solutions for Section 11.2
EXERCISES
1. We can rewrite the left-hand side as 24 = 22·2 = (22 )2 = 42 . Thus the equation takes the form 42 = 4x , so x = 2.
Now let’s confirm using logarithms. Taking logs of both sides gives us
log(24 ) = log 4x
4 log 2 = x log 4
4 log 2/ log 4 = x
2 = x.
This confirms our previous result.
5. We can rewrite the left-hand side, noticing that 43 = 64. This leads us to 643 = (43 )3 = 43·3 = 49 . Therefore x = 9.
Now let’s confirm using logarithms. Taking logs of both sides gives us
log 643 = log 4x
3 log 64 = x log 4
3 log 64/ log 4 = x
9 = x.
This confirms our previous result.
9. We have
y
5
2
=
3
7
log(2/3)y = log(5/7)
y · log(2/3) = log(5/7)
log(5/7)
y=
log(2/3)
= 0.830.
13. We have
2p = 90
log 2p = log 90
p log 2 = log 90
log 90
p=
log 2
= 6.492.
11.2 SOLUTIONS
17. We have
5 · 10x = 35
35
= 7,
10x =
5
so
x = log 7 = 0.845.
21. We have
100(1.041)t = 520
520
1.041t =
100
log 1.041t = log(520/100)
t log 1.041 = log(520/100)
log(5.2)
t=
log 1.041
= 41.030.
25. We have
8(2/3)t = 20
(2/3)t = 2.5
t log(2/3) = log 2.5
log 2.5
t=
log(2/3)
= −2.260.
29. We have
z/17
2
3
2 z/17
3
z/17 2
log
3
z
2
· log
17
3
120
= 15
=
15
= 0.125
120
= log 0.125
= log 0.125
17 log 0.125
log 23
= 87.185.
z=
33. We have
820(1.031)t = 1140(1.029)t
1.031t
= 1140/820
1.029t
t
1.031
= 1140/820
1.029
117
118
Chapter Eleven /SOLUTIONS
t log(1.031/1.029) = log(1140/820)
log(114/82)
t=
log(1.031/1.029)
= 169.682.
37. We have
500(1.032)x = 750
1.032x = 750/500
x log 1.032 = log 1.5
log 1.5
x=
log 1.032
= 12.872.
41. Using, if log x = y then x = 10y , we have
log x = −3
x = 10−3
1
x=
= 0.001.
1000
45. We have
7 − 2 log x = 10
−2 log x = 3
log x = −1.5
x = 10−1.5
= 0.0316.
PROBLEMS
49. Since the x is not in the exponent, we do not expect x to be represented by a logarithm. To confirm, we solve:
102 x = 103 + 102
100x = 1000 + 100 = 1100
x = 11.
53. Since a = log x, we know that a is an exponent, and we do not expect x to involve logarithms. In fact, since the equation
involves log x, we expect the answer to involve exponentials. To confirm, we solve using the definition of a logarithm:
a = log x
means
x = 10a .
57. The function f is exponential, and it describes a city that begins with 1800 people and that grows by 5.5% each year.
Solving the equation gives
1800(1.055)t = 2500
2500
1.055t =
1800
11.3 SOLUTIONS
119
t log 1.055 = log(25/18)
log(25/18)
t=
log 1.055
= 6.136.
This tells us that the population reaches 2500 after a little more than 6 years
61. The function v is exponential. It tells us that initially, 80 thousand cubic feet per day of gas are delivered, and that this
rate is halved every 5 months. Solving gives
80 · 2−t/5 = 5
2−t/5 = 0.0625
t
− log 2 = log 0.0625
5
log 0.0625
t = −5 ·
log 2
= 20.
Alternatively, we can rewrite 80 · 2−t/5 = 5 as 2t/5 = 80/5 = 16 = 24 , from which we see t/5 = 4, so t = 20.
This tells us that the rate of production drops to 5 thousand cubic feet per day after 20 months.
Solutions for Section 11.3
EXERCISES
1. The quantity Q is increasing. To find the doubling time we can solve as follows:
200(1.21)t = 400
1.21t = 2
log 1.21t = log 2
t log 1.21 = log 2
log 2
t=
log 1.21
= 3.636.
5. Let t be time in days since there was 10 kg. Since the quantity, A, is decaying exponentially, we have A = 10bt .
We know that at the end of 14 days, 95% of the mass, or 9.5 kg remains. Thus
9.5 = 10b14
Solving for b gives
b14 =
9.5
= 0.95
10
b = (0.95)1/14 = 0.99634.
Thus
A = 10(0.99634)t .
We want to solve for t making A = 1:
1 = 10(0.99634)t .
120
Chapter Eleven /SOLUTIONS
Dividing by 10 and taking logs gives
1
= (0.99634)t
10
1
= log(0.99634)t .
log
10
The properties of logs gives, to the nearest day,
−1 = t log(0.99634)
t=
−1
= 628 days.
log(0.99634)
9. The number, N , of SUV sales t years after 2003 will be
N = abt ,
where a is the number of sales in 2003 and b is the growth factor. We have b = 1.30, so
N = a(1.30)t .
We want to know the time at which N = 2a, so we solve
2a = a(1.30)t ,
for t. Thus,
(1.30)t = 2,
so
t log 1.30 = log 2,
giving
log 2
= 2.64.
log 1.30
So the number of SUVs sold will double every 2.64 years.
t=
13. We have
1900(1.061)t = 5000
1.061t = 5000/1900
t log 1.061 = log(5000/1900)
log(5000/1900)
t=
log 1.061
= 16.3411.
The investment will be worth $5000 in a little more than 16 years.
PROBLEMS
17. You start with $10,000, and your money grows at a rate of 14% per year. Therefore, after n years, you have 10,000(1.14)n
dollars in your account.
(a) We first need to solve 15,000 = 10,000(1.14)n for n. Divide both sides by 10,000, and then take log of both sides.
This gives us
log(3/2) = log(1.14)n
log(3/2) = n log 1.14
n = log(3/2)/ log 1.14 = 3.094.
Thus in just over 3 years, you have $15,000 in your account.
11.3 SOLUTIONS
121
(b) To find when you have $20,000, we perform similar steps. We start with 20,000 = 10,000(1.14)n ; divide both sides
by 10,000 and take log of both sides. This leads us to
log 2 = log(1.14)n
log 2 = n log 1.14
n = log 2/ log 1.14 = 5.290.
So after a little over 5 years you have $20,000 in the account. Note that it took about 3 years to earn the first $5,000,
and then only about another 2.3 years to earn the next $5,000.
21. (a) The volume of the balloon doubles each minute; if we let V0 be the initial volume of the balloon, we have V = V0 2t ,
where t is time in minutes. We began measuring the volume of the balloon at time t = 0, and when t = 10 we have
V = 512. Thus
512 = V0 210 .
10
Divide both sides by 2 = 1024 to solve for V0 :
V0 =
1
512
= .
1024
2
Thus the initial volume is 1/2 cm3 .
(b) Since the volume doubles every minute, it was exactly half the volume reached at 10 minutes at 9 minutes; hence a
volume of 256 cm3 at 9 minutes gets doubled to 512 cm3 at 10 minutes.
(c) To solve for t in the equation
100 = (1/2)(2t ),
multiply both sides by 2, and then take logs of both sides. This gives
log 200 = log 2t .
By properties of logs, we have
log 200 = t log 2,
and hence
t = log 200/ log 2 = 7.644 minutes.
25. (a) After t days,
Quantity remaining = 5bt ,
where b is the factor by which the quantity decays each day. We know that when t = 8, half the original quantity, or
2.5 mg, remains, so
2.5 = 5 · b8
b8 = 0.5
b = (0.5)1/8 = 0.917.
Thus, at time t
After a week, or 7 days,
Quantity remaining = 5(0.917)t .
Quantity remaining = 5(0.917)7 = 2.726 mg.
(b) Only 1 mg remains when
1 = 5(0.917)t .
Solving for t by taking logarithms,
log 1 = log 5 + log (0.917)t
0 = log 5 + t log(0.917)
log 5
= 18.574 days.
t=−
log(0.917)
29. We want t so that 2a = abt , so 2 = bt , log 2 = t log b, giving t =
log 2
log b
hours.
122
Chapter Eleven /SOLUTIONS
Solutions for Section 11.4
EXERCISES
1. We have
8e−0.5t = 3
e−0.5t = 0.375
divide by 8
−0.5t = ln 0.375
definition of natural log
t = −2 ln 0.375.
An approximate value is t = 1.962. Checking our answer, we see that
8e−0.5(1.962) = 8e−0.981
= 2.999,
as required (taking rounding error into account).
5. We have
23et/90 = 1700
1700
divide by 23
et/90 =
23
t
1700
= ln
90
23
1700
.
t = 90 ln
23
An approximate value is t = 387.26. Checking our answer, we see that
23et/90 = 23e387.26/90
= 23e4.3029
= 1700.018,
as required (taking rounding error into account).
9. We have:
(a)
3000(0.926)t = 600
(b) 3000(0.926)t = 600
(0.926)t = 0.2
0.926t = 0.2
t
log 0.926 = log 0.2
ln 0.926t = ln 0.2
t · log 0.926 = log 0.2
log 0.2
t=
log 0.926
= 20.934.
t · ln 0.926 = ln 0.2
ln 0.2
t=
ln 0.926
= 20.934.
dividing both sides by 3000
applying (a) log, (b) ln
In this case, it makes little difference whether we use log or ln, since the equation involves neither base 10 nor base e. To
verify our solutions, we see that
3000(0.926)t = 3000(0.926)20.934
= 600.006,
as required (taking rounding error into account).
11.4 SOLUTIONS
123
13. We have Q = 350(1.318)t , an exponential function of the form Q = abt :
•
•
•
•
The starting value is a = 350;
the growth factor is b = 1.318;
the growth rate is r = b − 1 = 0.318 = 31.8%; and,
the continuous growth rate is k = ln b = ln 1.318 = 0.2761 = 27.61%.
17. After t years, the value of this investment is given by V = 2500(1.082)t . To find when the value reaches $12,000, we
solve
2500(1.082)t = 12,000
12,000
1.082t =
2500
log 1.082t = log 4.8
t · log 1.082 = log 4.8
log 4.8
t=
log 1.082
= 19.903.
To check our answer, we find V at t = 19.903:
V = 2500(1.082)19.03 = 11,99.56,
which is as we should expect (taking rounding error into account).
21. We have log2 0.25 = −2 because 2−2 = 1/4 = 0.25.
25. We have log20 400 = 2 because 202 = 400.
29. We have log100 10,000 = 2 because 1002 = 10,000.
33. We have
To verify our answer, we find
log 0.75
= log5 0.75
log 5
log 0.75
log 5
so b = 5, x = 0.75
= −0.179 and show that −0.179 is the exponent of 5 that gives 0.75:
5−0.179 = 0.750,
as required (taking rounding error into account).
PROBLEMS
37. We have:
f (200) = 90e200/499.7
= 134.296.
This tells us that 200 days after October 1, 2002, Wikipedia’s English-language site had about 134,000 articles.
41. We have
so
3 in ·
25.4 mm
= 76.2 mm
1 in
76.2 mm
φ = − log2
1 mm
log 76.2
=−
log 2
= −6.252.
124
Chapter Eleven /SOLUTIONS
45. We have
log 17
= 0.5 log 17
2
= log 170.5
√
= log 17,
so x =
√
17. To verify our answer, we must show that log 17/2 = 0.6152 is the exponent of 10 that gives
100.6152 = 4.123,
as required (taking rounding error into account).
49. Notice that, much as
log x
= log 5 x,
log 5
it is also true that
log3 x
= log5 x.
log3 5
To see this, write
5y = x
y is exponent of 5 that yields x
y
log3 (5 ) = log3 x
y log3 5 = log3 x
log3 x
y=
.
log3 5
This means we can write
log3 81
4
=
Because 34 = 81
log3 5
log3 5
= log5 81,
so x = 81. To verify our answer, we must show that
4
4
=
= 2.730
log3 5
log 5/log 3
is the exponent of 5 that gives 81:
52.73 = 80.945,
as required (taking rounding error into account).
Solutions for Chapter 11 Review
EXERCISES
1. We have log 100 = 2 because 102 = 100.
5. Since 10 to any real-valued power is positive, log(−1) is undefined.
9. Using log(10N ) = N, we have log 103.68 = 3.68.
13. Since 10 to any real-valued power is positive, log(−10) is undefined. Thus, 10log(−10) is also undefined.
17. We use the fact that if y = log x then 10y = x. So log 10 = 1 means that 101 = 10.
√
17 = 4.123:
SOLUTIONS to Review Problems for Chapter Eleven
125
21. We use the fact that if y = log x then 10y = x. So log w = r means that 10r = w.
25. We use the fact that if 10y = x, then y = log x. So 10−1 = 0.1 means that log 0.1 = −1.
29. (a) If 10y = x, then y = log x. So 102 = 100 means
√ log 100 = 2.
√
(b) If x2 = y, then y = x. So 102 = 100 means 100 = 10.
33. Using the log rule log N p = p · log N , we let N = 1000 and p = 2x3 :
log 10002x
3
= 2x3 log 1000
= 2x3 · 3
= 6x3 .
37.
log
√
1000x = log
103
x 1/2 = log 103x/2
=
3x
.
2
41. Start by dividing both sides by 8; then take logs of both sides, giving
log(1.5t ) = log(60.75/8)
t log 1.5 = log(60.75/8)
t = log(60.75/8)/ log 1.5
t = 5.
45. Divide both sides by 16 to get (1/2)t = 1/4. You may immediately realize that (1/2)2 = 1/4; in case you don’t see this,
we can always take logs of both sides.
t
1
4
1
1
t log = log
2
4
log
1
2
= log
t = log
t = 2.
49. Dividing by 5, we get
6x =
1
1
/ log
4
2
2
= 0.4.
5
Taking logs of both sides we get
log 6x = log 0.4.
Using the properties, we get
x log 6 = log 0.4.
Solving for x, we get
x=
log 0.4
= −0.511.
log 6
126
Chapter Eleven /SOLUTIONS
53. Start by dividing both sides by 4, then take logs of both sides, giving
log(3t
2
−7t+19
) = log 39 .
We could simplify 39 as 19,683, but we could also note that this previous equation says that 3 raised to some power is the
same as 3 to the 9th power. Therefore the “some power” must equal 9; this tells us that t2 − 7t + 19 = 9. However, we’ll
proceed as usual and we’ll see that we get the same result.
log(3t
2
−7t+19
) = log 39
(t2 − 7t + 19) log 3 = 9 log 3
t2 − 7t + 19 = 9.
This gives us the same quadratic equation as before, which we must now solve. Subtract 9 from each side, giving
t2 − 7t + 10 = 0. This we can factor; (t − 5)(t − 2) = 0. Therefore we have two solutions; t = 2 and t = 5.
57. Dividing by 10(0.6)t , we get
0.4 =
(0.6)3t
= (0.6)3t−t = (0.6)2t .
(0.6)t
Taking logs of both sides we get
log 0.4 = log(0.6)2t .
Using the properties, we get
log 0.4 = 2t log 0.6.
Solving for t, we get
t=
log 0.4
.
2 log 0.6
61.
1
100
1
√
3
102
1
102/3
10−2/3
2
− .
3
10x = √
3
=
=
=
x=
65.
x
10x = 3 · 10x − 3
10 − 3 · 10x = −3
−2 · 10x = −3
10x = 1.5
x = log 1.5.
69.
30 1 − 0.5t = 25
1 − 0.5t =
25
30
SOLUTIONS to Review Problems for Chapter Eleven
127
5
−1
6
1
=
6
1
= log
6
1
= log
6
log 16
=
.
log 0.5
−0.5t =
0.5t
log 0.5t
t log 0.5
t
73. Taking logs of both sides we get
log 250(0.8)t = log 800(1.1)t .
Using the properties, we get
log 250 + t log 0.8 = log 800 + t log 1.1.
Collecting the terms with t, we get
t log 0.8 − t log 1.1 = log 800 − log 250.
Factoring out t, we get
Solving for t, we get
t(log 0.8 − log 1.1) = log 800 − log 250.
t=
log 800 − log 250
.
log 0.8 − log 1.1
77. Here is one approach:
2 log x = log 9
1
log x = · log 9
2
log x = log 91/2
log x = log 3
x = 3.
Here is another approach:
2 log x = log 9
log(x2 ) = log 9
x2 = 9.
81.
At this step, we must remember that x > 0, for otherwise the original equation would not make sense: 2 log x is not
defined if x ≤ 0. Thus, taking square roots, we do not get ±3, we get x = 3.
1
3
√
3
x = 101/3 = 10.
log x =
128
Chapter Eleven /SOLUTIONS
85.
2 log x + 1 = 5 − log x
2 log x + log x = 4
3 log x = 4
4
log x =
3
x = 104/3 = 21.544.
89. The equation is equivalent to e0 = 1.
93. The equation is equivalent to e4 = 7r.
97. The equation is equivalent to ln 0.018 = −4.
101. We have ln(ln e) = ln(1) = 0.
105. We have
ln
√
e−
√
√
ln e = ln e1/2 − ln e
√
1
= ln e − ln e
2
1 √
= − 1
2
1
=− .
2
109. We have 3 ln x−2 + ln x6 = 3(−2) ln x + 6 ln x = 0.
113. The expression cannot be simplified because it is the natural logarithm of a sum that cannot be simplified.
117. Rewrite the expression inside the log so that it is a power of the base. Since
1
25
= 5−2 , then log5
log 5
= 1.465
log 3
ln 5
=
1.465
ln 3
121. (a) log 3 5 =
(b) log 3 5 =
The values are the same.
125. log 2 3 =
log 3
log 2
= 1.585 Note that using the natural logarithm would yield the same value.
PROBLEMS
129. We use log at = t log a, giving
1/2
1
10 2 log 5 = 10log 5
,
and then 10log A = A, so
1/2
1
10 2 log 5 = 10log 5
= 51/2 =
√
5.
133. (III). By the definition of a logarithm: log x = b.
137. (IV). By the definition of logarithms, if bx = b1/10 , then
x = log(b1/10 ) =
1
log b.
10
141.
√ √
log 2xz 3 y = log 2 + log x + log z 3 + log y
= log 2 + log x + 3 log z + log y 1/2
1
25
= −2
SOLUTIONS to Review Problems for Chapter Eleven
129
= log 2 + log x + 3 log z + 0.5 log y
= log 2 + u + 3w + 0.5v.
145. Since 10log N = N we have
10log(2x+1) = 2x + 1.
149. We have
50 · 3t/4 = 175
175
3t/4 =
= 3.5
50
log 3t/4 = log 3.5
t
log 3 = log 3.5
4
t
log 3.5
=
4
log 3
4 log 3.5
.
t=
log 3
153. We have
2 · 102x+1 = 25
25
102x+1 =
= 12.5
2
2x + 1 = log 12.5 = 1.0969
1.0969 − 1
= 0.485.
x=
2
157.
2 log(x) − 1 = 5
2 log x = 6
log x = 3
x = 103
= 1000.
161.
log(log x) = 1
10log(log x) = 101
log x = 10
10log x = 1010
x = 10,000,000,000.
165. If L is the number of leakages now, then after one year, there will be 1.03L leakages, after two years (1.03)2 L leakages,
and after n years (1.03)n L leakages. To find n for which the number leaking is twice the current number, we need to
solve (1.03)n L = 2L for n, namely, n = log 2/log 1.03 = 23.449 years.