Exercises 4.1
173
Chapter 4: Exponential and Logarithmic Functions
EXERCISES 4.1
1.
3.
a.
e2 ≈ 7.389
a.
e3 ≈ 20.086
b.
e −2 ≈ 0.135
b.
e −3 ≈ 0.050
c.
e1/ 2 ≈ 1.649
c.
e1/ 3 ≈ 1.396
a.
e5e −2 = e5+( −2) = e3
a.
e −3e5 = e( −3) +5 = e2
b.
e5 = e5−3 = e2
e3
b.
e 4 = e 4 = e 4−1 = e3
e e1
c.
e5e −1 = e5e −1 = e5+( −1) = e4 = e4−( −1) = e5
e −2 e e −2 e1
e −2+1
e −1
c.
2.
4.
5.
6.
7.
8.
e2 e = e2 e1 = e2+1 = e3
e e
e −2 e −1 e −2+( −1) e −3
= e3−( −3) = e6
−2 −1
9.
5.697
10.
0.914
11.
a.
12.
a.
on [0, 10] by [0, 10]
0.01 x
x is higher than e
.
on [0, 5] by [0, 20]
x
2
e is higher than x .
b.
b.
on [0, 6] by [0, 200]
x
3
e is higher than x for large values of x.
on [0, 1000] by [0, 1000]
0.01 x
e
is higher than x when x is near 1000.
© 2010 Brooks/Cole, Cengage Learning.
174
Chapter 4: Exponential and Logarithmic Functions
c.
on [0, 10] by [0, 10,000]
x
4
e is higher than x for large values of x.
d.
e.
13.
a.
on [0, 15] by [0, 1,000,000]
e x is higher than x 5 for large values of x.
ex will exceed any power of x for large
enough values of x.
For n = 1 m = 1 (annual compounding),
( )
14.
For n = 1 m = 1 (annual compounding),
a.
( )
nt
P 1+ r
simplifies to P(1 + r)t .
n
When P = 1000, r = 0.1, and t = 8 , the value
is
1000(1 + 0.1)8 = 1000(1.1)8 = 2143.59
The value is $2143.59.
b. For quarterly compounding,
n = 4, P = 1000, r = 0.1, and t = 8 . Thus,
1000(1 + 0.12)8 = 1000(1.12)8 = 2475.96
The value is $2475.96.
b. For quarterly compounding,
n = 4, P = 1000, r = 0.12, and t = 8 . Thus,
(
For continuous compounding,
P = 1000, r = 0.10, and t = 8 . Thus,
= 1000(1.03)32
= 2575.08
The value is $2575.08.
For continuous compounding,
P = 1000, r = 0.12, and t = 8 . Thus,
c.
P = 100, r = 0.02, and n = 3 , compounded
weekly, which is 52 times per year. This
gives a value of
100(1 + 0.02)52⋅3 = 100(1.02)52⋅3
= 100(1.02)156
= $2196
b. The “vig” is equal to the amount owed after
three years minus the amount loaned. This is
$2196 − $100.00 = $2096.
© 2010 Brooks/Cole, Cengage Learning.
= 1000(1 + 0.03) 4⋅8
Pe rt = 1000e0.12 ⋅8 − 1000e0.96 = 2611.70
The value is $2611.70.
Pe rt = 1000e0.1⋅8 − 1000e0.8 = 2225.54
The value is $2225.54.
a.
4⋅8
1000 1 + 0.12
4
= 1000(1.025) = 2203.76
The value is $2203.76.
15.
)
1000(1 + 0.1
)4⋅8 = 1000(1 + 0.025) 4⋅8
4
32
c.
nt
P 1+ r
simplifies to P(1 + r)t .
n
When P = 1000, r = 0.12, and t = 8 , the
value is
16.
For quarterly compounding, r = 0.05 and
n = 2010 – 1626 = 384.
(
24 1 + 0.05
4
)
4⋅384
= 24(1.0125)4⋅384
= 24(1.0125)1536
= $4,644,941,190
Exercises 4.1
17.
175
The stated rate of 9.25% (compounded daily) is
the nominal rate of interest. To determine the
effective rate of interest, use the compound
interest formula, P (1 + r )n , with r =
18.
9.25%
number of days
To determine the effective rate of interest, first
we determine r in the cases of 360 days and 365
and n = 360.
days. For 360 days, r = 0.07
360
360
P (1 + r )n = P ⎛⎜1 + 0.07 ⎞⎟
≈ 1.0725P
360 ⎠
⎝
The effective rate is 7.25%.
and n = 365.
For 365 days, r = 0.07
360
and n = number of days in a year. Since some
banks use 365 days and some use 360 in a year,
we will try both ways.
If n = 365 days, then
r = 9.25% = 0.0925 ≈ 0.0002534
365
365
365
P (1 + r )n = P ⎛⎜1 + 0.07 ⎞⎟
≈ 1.0725P
365 ⎠
⎝
The effective rate is 7.25%. The error in the
annual yield should be 7.25%.
Then P (1 + r )n = P (1.0002534)365 ≈ 1.0969 P .
Subtracting 1 gives 0.0969, which expressed as a
percent gives the effective rate of interest as
9.69%.
If n = 360 days, then r = 0.0925 ≈ 0.0002569 .
360
Then P (1 + r )n = P (1.0002569)360 ≈ 1.0969 P
and the effective rate is also 9.69%.
Thus, the error in the advertisement is 9.825%.
The annual yield should be 9.69% (based on the
nominal rate of 9.25%).
19.
If the amount of money P invested at 8%
compounded quarterly yields $1,000,000 in
= 0.02 and
60 years, then r = 0.08
4
20.
The yield is $65,000, r =
0.06
4
= 0.0152 , and
n = 18 · 4 = 72.
65,000 = P (1 + 0.015)72
65,000
P=
≈ $22, 251
(1 + 0.015)72
n = 60 · 4 = 240.
1, 000, 000 = P(1 + 0.02)240
1, 000, 000
P=
≈ $8629
(1 + 0.02) 240
21.
For compounding annually, r = 0.06 and n = 45.
1000
Present value = P =
≈ $72.65
n
(1 + r )
(1 + 0.06)45
22.
For 6% compounded annually, r = 0.06 and
n = 4.
Present value = P n = 1000 4 ≈ $792.09
(1 + r )
(1 + 0.06)
23.
For 10% compounded annually, r = 0.10 and
n = 3.
Present value = P n = 1000 3 ≈ $751.31
(1 + r )
(1 + 0.10)
24.
For 11% compounded annually, r = 0.11 and
n = 10.
P (1 + r )n = 2.35(1 + 0.11)10 ≈ $6.67 million
© 2010 Brooks/Cole, Cengage Learning.
176
25.
Chapter 4: Exponential and Logarithmic Functions
To compare two interest rates that are
compounded differently, convert them both to
annual yields.
10% compounded quarterly:
26.
4
P (1 + r )n = P ⎛⎜1 + 0.08 ⎞⎟ ≈ P(1.0824)
⎝
4 ⎠
The effective rate of interest is 8.24%.
7.8% compounded continuously:
P (1 + r )n = P (1.025) 4 ≈ P(1.1038)
Subtracting 1, 1.1038 – 1 = 0.1038
The effective rate of interest is 10.38%.
9.8% compounded continuously:
Pern = Pe0.078 ≈ P (1.0811)
The effective rate of interest is 8.11%.
Thus 8% compounded quarterly (an effective rate
of 8.24%) is better than 7.8% compounded
continuously (an effective rate of 8.11%).
Pern = Pe0.098 ≈ P (1.1030)
Subtracting 1: 1.1030 – 1 = 0.1030
The effective rate of interest is 10.30%.
Thus, 10% compounded quarterly is better than
9.8% compounded continuously.
27.
Since the depreciation is 35% per year, r = –0.35.
a.
29.
n
28.
4
P (1 + r ) = 15, 450(1 − 0.35) ≈ $2758
b. P (1 + r )n = 15, 450(1 − 0.35)0.5 ≈ $12, 456
Since 2015 is 10 years after 2005, x = 10.
8% compounded quarterly:
For 12.6% compounded annually, r = 0.126 and
n = 10.
P (1 + r )n = 37(1 + 0.126)10 ≈ $121 million
30.
6.44e0.0123(10) ≈ 7.28 billion
The population of China is given by
1.3(1.006)t billion, where t is the number of
years after 2005.
In 2025 (t = 20), the population of China will be
1.3(1.006) 20 ≈ 1.47 billion .
The population of India is given by 1.13(1.015)t
billion, where t is the number of years after 2005.
In 2025 (t = 20), the population of India will be
1.13(1.015) 20 ≈ 1.52 billion .
India's population will be larger in 2025.
31.
a.
1 − (0.9997)100(25) ≈ 0.5277
32.
If a mosquito breeds 300 mosquitoes on average,
then the 300 children will breed
(300)(300) = 3002 = 90, 000 grandchildren, and
the 90,000 grandchildren will breed
3
(90,000)(300) = 300
= 27,000,000 great grandchildren.
b.
33.
1 − (0.9997)100(40) ≈ 0.6989
The proportion of light that penetrates to a depth
34.
of x feet is given by e −0.44 x .
a.
a.
f (24) = 400e−0.012(24) ≈ 300 mg
b.
f (48) = 400e−0.012(48) ≈ 225 mg
If the depth is 3 feet,
e−0.44 x = e−0.44(3) = e−1.32 ≈ 0.267 or 26.7%
b. If the depth is 10 feet,
e −0.44 x = e −0.44(10) = e−4.4 ≈ 0.0123 or
1.23%
35.
a.
f (24) = 2e −0.018(24) ≈ 1.3 mg
b.
f (48) = 2e −0.018(48) ≈ 0.8 mg
© 2010 Brooks/Cole, Cengage Learning.
36.
If the colony doubles every hour, the dish
was 50% covered at 11 A.M.
b. If the dish was 50% covered at 11 A.M., it
was 25 = 12 (50%) covered at 10 A.M.
a.
Exercises 4.1
177
37.
S (10) = 100 + 800e −0.2(10) ≈ 208
39.
a.
After 15 minutes, t = 15
=
60
1
4
38.
= 0.25 and
40.
p(10) = e −0.01(10) ≈ 0.90 or 90%
a.
p(0) = 100 1 + e = 100%
1 + e0+1
b.
p(2) = 100 1 + e ≈ 17.6%
1 + e 2+1
T (0.25) = 70 + 130e −1.8(0.25) ≈ 153 degrees
b. After 30 minutes, t =
30
60
= 0.5 and
T (0.5) = 70 + 130e −1.8(0.5) ≈ 123 degrees
41.
If S = 400, x = 10, and r = 0.01, then the Reed-
42.
− (0.01)(10)
805e0.062(10) ≈ $1504 thousand
Frost model is I = 400(1 − e
) ≈ 38 .
The model estimates there will be about 38 newly
infected people.
43.
We have P = $8000, n = 4, and the value after 4
years is $10,291.73.
44.
The air pressure is (1 − 0.004)a /100 percent of the
original pressure, where a is the change in
altitude.
(
7347−30
100
10,380.65 = 9000 ⎛⎜1 + r ⎞⎟
⎝ 4⎠
⎛ 10,380.65 ⎞
r = ⎜8
− 1⎟ ⋅ 4 ≈ 0.072 = 7.2%
9000
⎝
⎠
46.
)
(1 − 0.004)
≈ .75 = 75%
So the air pressure decreased by about
100 – 75 = 25%.
47.
a.
48.
on [0, 100] by [0, 100]
b. During the year 2087 ( x ≈ 82.08)
49.
$40, 000e0.195(2) ≈ $59,079
We have P = $9000, n = 2 · 4 = 8, and the value
after 2 years is $10,380.65.
8
10, 291.73 = 8000(1 + r )4
10, 291.73
4
= 1+ r
8000
10, 291.73
r=4
− 1 ≈ 0.065 = 6.5%
8000
45.
Since 2010 is 10 years after 2000, x = 10.
(1 − 0.025)t percent will still be present after t
years.
a.
(1 − 0.025)50 ≈ 0.28 = 28%
b.
(1 − 0.025)100 ≈ 0.08 = 8%
a.
on [–400, 400] by [0, 700]
b. 630 feet
50.
It is not possible to evaluate f ( x ) = a x with a
negative base (a < 0) and x = 0.5, because the
function is not defined for all values of x.
For example, if a = –3, and x = 0.5,
f (0.5) = ( −3)0.5 = −3 , which is not a real
number.
51.
lim e x = ∞ and
x →∞
lim e x = 0 . The x-axis
x →−∞
(y = 0) is a horizontal asymptote.
52.
lim e − x = 0 and
x →∞
lim e − x = ∞ .
x →−∞
The x-axis (y = 0) is a horizontal asymptote.
© 2010 Brooks/Cole, Cengage Learning.
178
Chapter 4: Exponential and Logarithmic Functions
For very large values of x, e x will be largest
because it will have the largest exponent.
54.
55.
That its growth is proportional to its size.
56.
Semiannually, quarterly, daily, continuously,
because the benefit to the depositor increases as
the interest is compounded more often.
57.
With 5% monthly, you must wait until the end of
the month to receive interest, while for 5%
compounded continuously you begin receiving
interest right away, so the interest begins earning
interest without any delay.
58.
Daily compounding gives less profit than
continuous compounding. The banker was
incorrect.
59.
Depreciation by a fixed percentage gives the
bigger decrease in the first year, and the straight
line gives the bigger decrease in the last year.
60.
The continuously compounded rate is less than
10%, because 10% more at the end of the year
accounts for the interest on the interest as well as
the interest on the principle.
2.
a.
log3 27 = log3 33 = 3
53.
e x is a continually increasing function, so if
x < y, then e x < e y .
EXERCISES 4.2
1.
3.
a.
log5 25 = log5 52 = 2
b.
log3 81 = log3 34 = 4
b.
log 2 16 = log 2 24 = 4
c.
log3 13 = log3 3−1 = −1
c.
log16 4 = log16 161/ 2 =
d.
log3 19 = log3 3−2 = −2
d.
log 4
e.
log 4 2 = log 4 41/ 2 =
e.
log 2 18 = log 2 2−3 = −3
f.
log 4
f.
log9 13 = log9 9 −1/ 2 = − 12
a.
ln( e10 ) = 10
a.
b.
ln e = ln e1/ 2 =
b.
ln( e −5 ) = −5
ln e = 1
c.
ln e4 = ln e4 / 3 =
c.
ln 3 e = ln e1/ 3 = 13
d.
ln e5 = ln e5 / 2 =
e.
ln
f.
ln(ln e) = ln 1 = 0
1
2
1
2
= log 4 4 −1/ 2 = − 12
3
4.
1
2
4
3
d. ln 1 = 0
because ln(ee ) = e
e.
ln(ln( ee )) = ln e
=1
f.
ln 13 = ln e −3 = −3
( )
e
5.
f ( x) = ln(9x) − ln 9 = ln 9 + ln x − ln 9
= ln x
6.
7.
f ( x) = ln(x 3 ) − ln x = 3 ln x − ln x
= 2 ln x
8.
9.
()
f ( x ) = ln x + ln 4 = ln x − ln 4 + ln 4
4
= ln x
© 2010 Brooks/Cole, Cengage Learning.
10.
1
4
1
2
= log 4 4 −1 = −1
5
2
( 1e ) = ln e−1 = −1
()
f ( x ) = ln x + ln 2 = ln x − ln 2 + ln 2
2
= ln x
f ( x) = ln(4 x) − ln 4 = ln 4 + ln x − ln 4
= ln x
f ( x) = ln(x 5 ) − 3 ln x = 5 ln x − 3 ln x
= 2 ln x
Exercises 4.2
179
11.
f ( x) = ln(e 5x ) − 2 x − ln1 = 5 x − 2 x − 0
= 3x
12.
f ( x) = ln(e −2x ) + 3 x + ln 1 = −2 x + 3x + 0
=x
13.
f ( x ) = 8 x − eln x = 8 x − x = 7 x
14.
f ( x ) = eln x + ln( e − x ) = x − x = 0
15.
The domain of ln( x 2 − 1) is the values of x such
16.
that x 2 − 1 > 0
x2 > 1
x > 1 or x < −1
The domain is {x x > 1 or x < −1} .
that 1 − x 2 > 0
− x 2 > −1
x2 < 1
−1 < x < 1
The domain is {x − 1 < x < 1} . The range is
The range is \ .
17.
a.
We use the formula P (1 + r )n with monthly
1 ⋅ 24% = 2% = 0.02 .
interest rate r = 12
Since double P dollars is 2P dollars, we
solve
P (1 + 0.02)n = 2 P
1.02n = 2
ln(1.02n ) = ln 2
n ln1.02 = ln 2
n = ln 2 ≈ 0.6931 ≈ 35
ln1.02 0.0198
Since n is in months, we divide by 12 to
convert to years.
A sum at 24% compounded monthly doubles
in about 2.9 years.
b. To find how many years it will take for the
investment to increase by 50%:
P (1 + r )n = 1.5P
(1 + r )n = 1.5
ln(1 + r )n = ln1.5
n ln(1 + r ) = ln1.5
Now, r = 1 ⋅ 24% = 2% = 0.02
12
thus, n = ln1.5 = ln1.5 = 0.4055 ≈ 20.5
ln(1 + r ) ln1.02 0.0198
Since n is in months, we divide by 12 to
convert to years.
20.5 ≈ 1.7 years
12
A sum at 24% compounded monthly
increases by 50% in about 1.7 years.
The domain of ln(1 − x 2 ) is the values of x such
18.
{y
y ≤ 0} .
a.
r = 0.36
= 0.03 . Since double P dollars is 2P
12
dollars, we solve
P (1 + 0.03)n = 2 P
1.03n = 2
ln(1.03n ) = ln 2
n = ln 2 ≈ 23.4 months
ln1.03
The money doubles in 23.4 ≈ 1.95 years
12
b. To find how many years it takes to increase
by 50%, which is 1.5P, we solve
1.5P = P (1 + 0.03)n
1.5 = 1.03n
ln1.5 = ln(1.03)n
n = ln1.5 ≈ 13.7 months
ln1.03
The money increases by 50% in
13.7 ≈ 1.14 years .
12
© 2010 Brooks/Cole, Cengage Learning.
180
19.
Chapter 4: Exponential and Logarithmic Functions
a.
rn
We use Pe with r = 0.07. Since triple P
dollars is 3P dollars, we solve
20.
Pe0.07n = 3P
e0.07n = 3
0.07n = ln 3
n = ln 3 ≈ 15.7 years
0.07
b. If P increases by 25%, the total amount is
1.25P. We solve
Pe0.06n = 1.75P
e0.06n = 1.75
n = ln1.75 ≈ 9.3 years
0.06
Pe
= 1.25P
0.07 n
e
= 1.25
0.07n = ln1.25
n = ln1.25 ≈ 3.2 years
0.07
If the depreciation is 30% = 0.3 per year, then
22.
P (1 + 0.029)n = 2 P
1.029n = 2
ln1.029n = ln 2
n ln1.029 = ln 2
n = ln 2 ≈ 24.2 years
ln1.029
24.
We use P(1 + r) n with r = 0.16, and we solve
r = –0.3. We use the interest formula P (1 + r )n
and we solve
P (1 − 0.3)n = 0.5P
0.7n = 0.5
ln 0.7n = ln 0.5
n = ln 0.5 ≈ 1.9 years
ln 0.7
23.
We use the interest formula P (1 + r )n with
r = 2.4% = 0.024. Since the increase is 50%, we
must find the number of years to reach 1.5P. We
solve
P (1 + 0.16)n = 2 P
1.16n = 2
n = ln 2 ≈ 4.7 years
ln1.16
P (1 + 0.024)n = 1.5P
1.024n = 1.5
n = ln1.5 ≈ 17.1 years
ln1.024
25.
We want to find the value of t that produces p(t)
= 0.9.
26.
0.99 = 1 − e−0.03t
e−0.03t = 0.01
t = ln 0.01 ≈ 154 days
−0.03
28.
900, 000 = 1, 000, 000(1 − e−0.4t )
0.9 = 1 − e−0.4t
−0.4t
e
= 0.1
t = ln 0.1 ≈ 5.8 hours
−0.4
0.9 = 1 − e−0.03t
e−0.03t = 0.1
−0.03t = ln 0.1
t = ln 0.1 ≈ 77 days
−0.03
27.
p(5) = 0.9 – 0.2 ln 5 ≈ 0.58 or 58%
© 2010 Brooks/Cole, Cengage Learning.
rn
We use Pe with r = 0.06. Since quadruple
P dollars is 4P dollars, we solve
Pe0.06n = 4 P
e0.06n = 4
0.06n = ln 4
n = ln 4 ≈ 23.1 years
0.06
b. A 75% increase gives 1.75P, and we solve
0.07 n
21.
a.
Exercises 4.2
29.
181
s (t ) = 100(1 − e−0.4t )
100(1 − e
) = 80
−0.4t
1− e
= 0.8
−0.4t
−e
= −0.2
−0.4t
e
= 0.2
−0.4t = ln 0.2
t = ln 0.2 ≈ 4 weeks
−0.4
30.
To find the number of years after which
2.3% = 0.023 of the original carbon 14 is left, we
solve
32.
−0.4t
31.
( )
The proportion of potassium 40 remaining after t
To estimate the age, we solve
e −0.00012t = 0.923
t = ln 0.923 ≈ 668 years
−0.00012
e −0.00012t = 0.023
−0.00012t = ln 0.023
t = ln 0.023 ≈ 31, 400 years
−0.00012
33.
5 = 20e−0.23t
1 = e−0.23t
4
⎛
1
ln ⎜ ⎞⎟ = −0.23t
⎝4⎠
ln 14
t=
≈ 6 hours
−0.23
34.
e−0.00054t = 0.9982
t = ln 0.9982 ≈ 3.3 million years
−0.00054
36.
Since rain forests are disappearing at an annual
rate of 1.8% = 0.018, then r = –0.018, and we
million years is e −0.00054t . If the skeleton
contained 99.91% of its original potassium 40,
then
e−0.00054t = 0.9991
ln e−0.00054t = ln 0.9991
−0.00054t = ln 0.9991
t = ln 0.9991 ≈ −0.0009004 ≈ 1.67
−0.00054
−0.00054
Therefore, the estimate of the age of the skeleton
of an early human ancestor discovered in Kenya
in 1984 is approximately 1.7 million years.
35.
To find when the radioactivity decreases to
0.001, we solve e −0.087t = 0.001
−0.087t = ln 0.001
t = ln 0.001 ≈ 79 days
−0.087
37.
1.16 x = 2
x = ln 2 ≈ 4.7 years
ln1.16
solve (1 − 0.018)t = 0.5
0.982t = 0.5
t = ln 0.5 ≈ 38 years
ln 0.982
38.
1.16 x = 1.5
ln(1.16 x ) = ln1.5
x ln1.16 = ln1.5
x = ln1.5 ≈ 2.7 years
ln1.16
© 2010 Brooks/Cole, Cengage Learning.
182
Chapter 4: Exponential and Logarithmic Functions
39.
40.
on [0, 60] by [0, 5]
a. 35 years
b. 55.5 years
41.
Let t = number of years. Since the rate is 6%
compounded quarterly, then r = 0.06 = 0.015
4
and the amount of money is
on [0, 24] by [0, 5]
a. 14.2 years
b. 22.5 years
42.
(1 + 0.015) 4t = 1.0154t .
on [0, 10] by [0, 3]
a. 9.9 years
b. 3.2 years
on [0, 16] by [0, 3]
a. 11.6 years
b. 6.8 years
43.
44.
on [0, 40] by [0, 1]
a. About 9 days
b. About 11 days
45.
on [0, 5] by [0, 6.3]
About every 4 days
46.
on [0, 7000] by [0, 1]
About 5300 years
47.
If the amount of radioactive waste is growing by
11.3% annually since 2000, then the amount of
waste x years after 2000 is 245, 000(1.113) x . We
must find the value of x such that (1.113) x = 2 .
1.113x = 2
ln1.113x = ln 2
x ln1.113 = ln 2
x = ln 2 ≈ 6.5
ln1.113
The amount will double in about 6.5 years.
© 2010 Brooks/Cole, Cengage Learning.
on [0, 250] by [0.8, 1]
About 228 million years
48.
We must find the value of x such that
(1.24) x = 2 .
1.24 x = 2
ln1.24 x = ln 2
x ln1.24 = ln 2
ln 2 ≈ 3.22 years
x = ln1.24
0.22 of a year is approximately 3 months, so the
doubling time for cell phone subscribers is 3 years
and 3 months.
Exercises 4.2
49.
We must find the value of x such that
183
50.
x
We must find the value of x such that
(1.0925) = 2 .
(1.0975) x = 2 .
1.0925x = 2
ln1.0925x = ln 2
x ln1.0925 = ln 2
ln 2 ≈ 7.83 years
x = ln1.0925
1.0975x = 2
ln1.0975x = ln 2
x ln1.0975 = ln 2
ln 2 ≈ 7.45 years
x = ln1.0975
0.83 of a year is approximately 10 months, so the
doubling time for the fifteen-year bond is 7 years
and 10 months.
0.45 of a year is approximately 6 months, so the
doubling time for the bond is 7 years and 6
months.
51.
e0 = 1
ln1 = 0
52.
ln( x ⋅ y ) = ln x + ln y
53.
ln x = ln x − ln y
y
54.
ln ( x n ) = n ⋅ ln x
55.
log 10 y or log y
56.
ln y
57.
If ln( −2) = x , then e x = −2 , but e x is always
positive. So ln( −2) is undefined.
58.
If ln 0 = x , then e x = 0 , so x would be −∞ ,
which is undefined.
59.
Continuous compounding would give the shortest
doubling time. Annual compounding would give
the longest doubling time.
60.
The drug with a larger absorption constant will
have more time between doses.
61.
a.
62.
N (t ) = Ke − ae for t ≥ 0, K > 0, b > 0.
N (0) = N 0 .
Substitute 0 for t.
e − kt = 1
2
− kt = ln 1
2
− kt = ln1 − ln 2
t = ln 2
k
The half-life is ln 2 .
k
b. If k = 0.018, half-life = ln 2 ≈ 39 hours .
0.018
− bt
N 0 = Ke − ae
− b (0)
0
= Ke − ae
= Ke − a
ln N 0 = ln Ke − a
ln N 0 = ln K + ln e − a
ln N 0 = ln K − a
a = ln K − ln N 0 = ln
(K N )
0
© 2010 Brooks/Cole, Cengage Learning.
184
Chapter 4: Exponential and Logarithmic Functions
(
63.
)
x
64.
f ( x) = H0 1 − 1
2N
Substitute 500 for N .
x
f ( x ) = H 0 ⎛⎜ 1 − 1 ⎞⎟ = H 0 (0.999) x
⎝ 2(500) ⎠
Reducing the frequency by 6%,
H 0 − 0.06 H 0 = H 0 (0.999) x
0.94 H 0 = H 0 (0.999) x
0.94 = (0.999) x
ln 0.94 = ln(0.999) x
ln 0.94 = x ln(0.999)
x = ln 0.94 ≈ 62 generations
ln 0.999
65.
67.
a.
For r = 6% , the doubling time is
approximately 72 = 12 years.
6
b.
P (1 + 0.06)n = 2 P
1.06n = 2
ln1.06n = ln 2
n = ln 2 ≈ 11.9 years
ln1.06
y = axe − bx for x ≥ 0, a > 1, b > 0.
Let x = y.
x = axe − bx
1 = ae − bx ( x ≠ 0)
1 = 1
bx
a
e
a = ebx
ln a = ln ebx
ln a = bx
x = ln a
b
x = y, so
y = ln a .
b
The equilibrium population is ln a .
b
66.
If the interest rate r is compounded
annually, the formula is
68.
P(1 + r )n = kP
(1 + r )n = k
ln(1 + r )n = ln k
n = ln k
ln(1 + r )
a.
For r = 1% , the doubling time is
approximately 72 = 72 years.
1
b.
P(1.01)n = 2 P
ln(1.01)n = ln 2
n = ln 2 ≈ 69.7 years
ln1.01
If the interest rate r is compounded
continuously, the formula is
Pe rn = kP
e rn = k
ln e rn = ln k
ln k
n=
r
EXERCISES 4.3
1.
d ( x 2 ln x ) = 2 x ln x + x 2
dx
x
= 2 x ln x + x
2.
d ⎛⎜ ln x ⎞⎟ =
dx ⎜⎝ x3 ⎟⎠
=
x3 ⎜ 1 ⎟ − 3 x 2 ln x
⎛
⎞
⎝ x⎠
3 2
(x )
1 − 3ln x
2
2
= x − 3x ln x
6
x
x4
3.
d ln x 2 = 2 x = 2
dx
x2 x
4.
d ln( x 3 + 1) = 3x 2
dx
x3 + 1
5.
1 −1 2
d ln x = d ln x1 2 = 2 x
= 1 x −1
2
dx
dx
x1 2
6.
d ln x = d (ln x )1 2 = 1 (ln x ) −1 2 1
2
dx
dx
x
© 2010 Brooks/Cole, Cengage Learning.
Exercises 4.3
185
7.
2
2
d ln( x 2 + 1)3 = 3( x + 1) (2 x ) = 6 x
2
dx
( x + 1)3
x2 + 1
8.
4
3
d ln( x 4 + 1)2 = 2( x + 1)(4 x ) = 8 x 3
4
2
dx
( x + 1)
x4 + 1
9.
d ln( − x ) = −1 = 1
−x x
dx
10.
d ln(5 x ) = 5 = 1
5x x
dx
11.
d ⎛ e x ⎞ = x 2 e x − 2 xe x = xe x − 2e x
dx ⎜⎝ x 2 ⎟⎠
( x2 )2
x3
12.
d ( x3e x ) = x3e x + 3 x 2 e x
dx
13.
d e x 3 + 2 x = e x 3 + 2 x (3x 2 + 2)
dx
14.
d 2e7 x = 2e7 x (7) = 14e7 x
dx
15.
d e x3
dx
16.
d ln(e x − 2 x) = e x − 2
dx
ex − 2x
3
= ex
( )=x e
3 3 3x2
2 x3 3
3
18.
d ( x ln x −
dx
19.
d ( x − e − x ) = 1 − e − x (−1) = 1 + e− x
dx
d ln e 2 x = d (2 x ) = 2 because ln e 2 x = 2 x
dx
dx
20.
d ln e x
dx
=
21.
d e1+ e x
dx
22.
d ln(e x
dx
+ e− x ) =
23.
d xe
dx
= ex e −1
24.
d ex
dx
=e
25.
d e3
dx
=0
26.
d (
dx
e) = 0
27.
d [ln( x 4
dx
28.
d [ x2e x
dx
17.
x d
x
= e1+ e dx
(1 + e x ) = e1+ e e x
because e3 is a constant
+ 1) − 4e x 2 − x ] =
4 x 3 − 4e x
x 4 +1
2
( 12 ) − 1 = x4 x+1 − 2e
3
x 2
4
− 2 ln x + ( x 2 + 1)3 ] = 2 xe x + x 2 e x − 2
( 1x ) + 3( x
2
x ) = (1)(ln x ) + x
d ( x)
dx
( 1x ) − 1 = ln x
= 1 because ln e x = x
e x + e− x ( −1)
x
e +e
−x
−x
x
= e x −e− x
e +e
because e is a constant
−1
+ 1)2 (2 x )
= 2 xe x + x 2 e x − 2x + 6 x ( x 2 + 1)2
29.
()
2
d ( x 2 ln x − 1 x 2 + e x + 5) = 2 x ln x + x 2 1 − 1 (2 x ) + e x2 (2 x ) + 0
2
dx
x 2
2
2
= 2 x ln x + x − x + 2 xe x = 2 x (ln x + e x )
()
30.
d ( e −2 x − x ln x + x − 7) = e −2 x ( −2) − ln x − x 1 + 1 + 0 = −2e −2 x − ln x
dx
x
31.
d (e x ln x 2 ) = e x ln x 2 + e x 2 = e x ln x 2 + 2 = 2e x ln x + 1
dx
x
x
x
32.
33.
) ( )
() (
d (e ln( x + 1)) = e ln( x + 1) + e
( x 1+ 1) = e (ln( x + 1) + x 1+ 1)
dx
x
x
x
x
d ⎛ ln 1 ⎞ = d ( ln x −2 ) = −2 d ( ln x ) = −2
dx ⎝ x 2 ⎠ dx
dx
x
© 2010 Brooks/Cole, Cengage Learning.
186
Chapter 4: Exponential and Logarithmic Functions
(
)
34.
d ⎛ ln 1 ⎞ = d ln e − x 2 = d ( − x 2 ) = −2 x
dx ⎜⎝ e x 2 ⎟⎠ dx
dx
35.
d ( e 2t + 1)3 = 3 ( e2t + 1) 2 ( 2e 2t ) = 6e 2t ( e2t + 1) 2
dt
36.
d ( e 2t + 4 )1/ 2 = 1 ( e 2t + 4 ) −1/ 2 ( 2e 2t ) =
dt
2
e 2t
e2t + 4
)
(
37.
d ( t 2 + 2 ln t )1/ 2 = 1 ( t 2 + 2 ln t ) −1/ 2 2t + 2 = t + 1/ t
dt
t
2
t 2 + 2 ln t
38.
d ( 2t + ln t ) 3 = 3 ( 2t + ln t ) 2 2 + 1
dt
t
39.
40.
( )
2
z
z
z 2
d ⎛ e z ⎞ = ( z − 1) e − e (2 z ) = e ( z − 2 z − 1)
⎜
⎟
2
dz ⎝ z 2 − 1⎠
( z 2 − 1)
( z 2 − 1)2
z z
z z
ez
d ⎛ e z ⎞ = (1 + e ) e − e e =
⎜
⎟
2
z
dz ⎝ 1 + e ⎠
(1 + e z )
(1 + e z )2
41.
d ⎛ 10 ⎞ = d 10 (1 + e −2 z ) −1 = −10 (1 + e −2 z ) −2 ( −2e −2 z ) = 20e −2 z
dz ⎝ 1 + e −2 z ⎠ dz
(1 + e−2 z )2
42.
d ⎛ 10 ⎞ = d 12 (1 + 2e − z ) −1 = −24 (1 + 2e − z ) −2 ( − e −2 z ) = 24e − z
dz ⎝ 1 + e −2 z ⎠ dz
(1 + 2e − z )2
43.
44.
x
−x
x
−x
x
−x
x
−x
2x
−2 x
2x
−2 x
−4
d ⎛ e x + e − x ⎞ = ( e − e )( e − e ) − ( e + e )( e + e ) = e − 2 + e − ( e + 2 + e ) =
⎜
⎟
x
−
x
2
2
dx ⎝ e − e ⎠
x
−
x
x
−
x
x
(e − e )
(e − e )
( e − e− x )2
x
−x
x
−x
x
−x
x
−x
2x
−2 x
2x
−2 x
4
d ⎛ e x − e − x ⎞ = ( e + e )( e + e ) − ( e − e )( e − e ) = e + 2 + e − ( e − 2 + e ) =
2
2
dx ⎜⎝ e x + e − x ⎟⎠
x
−
x
x
−
x
x
( e + e− x )2
(e + e )
(e + e )
45. a.
f ( x) =
f ′( x) =
47.
()
( x5 )2
46. a.
=
b.
f ′(1) = 1−ln1
= 1−10 = 1
6
a.
f ( x) = ln( x 4 + 48)
x 4 −5 x 4 ln x
x10
= 1 − 5ln x
x6
(1)
f ′( x) =
49.
ln x
x5
x5 1x −5 x 4 ln x
48.
4(2)3
32 = 32 = 1
16 + 48 64 2
a.
f ( x) = ln(e x − 3 x)
(2)4 + 48
=
50.
f ( x) = x 2 ln x − x 2
f ′( x) = 2 x ln x + x 2
b.
f ′(e) = 2e ln e − e = e
a.
f ( x ) = ln(e x + e − x )
f ′( x) = ex −3
e0 −3
e0 −3(0)
= 1 − 3 = −2
1− 0
© 2010 Brooks/Cole, Cengage Learning.
b.
3
3
f ′( x ) =
e −3 x
f ′(0) =
ln x + x3
a.
x
b.
3
f ′(1) = 4(1) ln 1+ (1) = 0 +1 = 1
4x
x 4 + 48
f ′(2) =
( 1x ) = 4 x
b.
3
b.
f ( x) = x 4 ln x
f ′( x) = 4 x3 ln x + x 4
f ′(0) =
e x + e− x ( −1)
e x + e− x
e 0 − e −0
e 0 + e −0
( 1x ) − 2 x = 2 x ln x − x
=
e x −e − x
e x + e− x
= 1−1 = 0
1+1
Exercises 4.3
51.
187
a.
f ( x ) = 5 x ln x
b.
f ′( x ) = 5ln x + 5 x 1 = 5ln x + 5
x
f ′(2) = 5ln 2 + 5
52.
()
a.
a.
b.
x
f ( x) = e
x
xe x − (1)e x xe x − e x
f ′( x ) =
=
x2
x2
3
3
3
f ′(3) = 3e −2e = 2e
9
(3)
54.
55.
f ′( x ) = e x
/2
( 22x ) = xe
a.
f ( x ) = ln( e x − 1)
x
f ′( x ) = xe
e −1
3
f ′(3) = 3e
e −1
x2 / 2
f ′(3) ≈ 1.052
d e − x 5 / 5 = e − x 5 / 5 ⎡ −5 ⎛ x 4 ⎞ ⎤ = − x 4 e − x 5 / 5
⎢ ⎜ 5 ⎟⎥
dx
⎣ ⎝ ⎠⎦
d 2 e − x5 / 5 = d ( − x 4 e − x5 / 5 )
dx
dx 2
57.
/2
2
f ′(2) = 2e 4 / 2 = 2e 2
f ′(2) ≈ 14.778
b.
f ′(3) ≈ 4.463
2
b.
f ′(2) ≈ 8.466
53.
f ( x) = e x
5
/5 ⎡
5
/5
= −4 x 3 e − x
5
/5
− x 4e− x
= −4 x 3 e − x
5
/5
+ x8e − x
56.
d 2 ( e − x 6 / 6 ) = −5 x 4 e − x 6 / 6
dx 2
⎛ x4 ⎞⎤
⎢5 ⎜ − 5 ⎟ ⎥
⎠⎦
⎣ ⎝
d ( ekx ) = ekx ( k ) = kekx
dx
d 2 ( ekx ) = d ( kekx ) = kekx ( k ) = k 2 ekx
dx
dx 2
d 3 ( ekx ) = d ( k 2 e kx ) = k 2 e kx ( k ) = k 3ekx
dx
dx 3
n
d ( ekx ) = k n e kx
dx n
59.
d ( e − x 6 / 6 ) = e − x 6 / 6 ⎡ 6 ⎛ − x 5 ⎞ ⎤ = − x 5e − x 6 / 6
⎢ ⎜ 6 ⎟⎥
dx
⎠⎦
⎣ ⎝
+ ( − x 5 )( e − x
= −5 x 4 e − x
58.
6
/6
6
/6
5⎞
⎛
) ⎜ −6 x ⎟
6 ⎠
⎝
+ x10 e − x
6
/6
d ( e − kx ) = e − kx ( −k ) = −ke − kx
dx
d 2 ( e − kx ) = −ke − kx ( −k ) = k 2 e − kx
dx 2
d 3 ( e − kx ) = k 2 e − kx ( −k ) = −k 3e − kx
dx 3
d n ( e − kx ) = ( −1)n k n e − kx
dx n
60.
on [–2, 2] by [–1, 2]
There is a relative maximum at (0, 1); no
relative minima. There are inflection points
at about (0.5, 0.61) and (–0.5, 0.61).
on [–3, 3] by [–1, 2]
There is a relative minimum at (0, 0); no
relative maxima. There are inflection points
at about (1, 0.39) and (–1, 0.39).
© 2010 Brooks/Cole, Cengage Learning.
188
Chapter 4: Exponential and Logarithmic Functions
61.
62.
on [–5, 5] by [–1, 4]
There is a relative minimum at (0, 0); no
relative maxima. There are inflection points
at about (1, 0.69) and (–1, 0.69).
63.
on [–5, 5] by [–1, 9]
There is a relative minimum at (0, 2); no
relative maxima. There are no inflection
points.
64.
on [–1, 8] by [–1, 3]
There is a relative maximum at about (2, 0.54);
relative minimum at (0, 0). There are inflection
points at about (0.59, 0.19) and (3.41, 0.38).
65.
on [–1, 5] by [–3, 1]
There is a relative maximum at about
(1, 0.37); no relative minimum. There is an
inflection point at about (2, 0.27).
66.
on [–2, 2] by [–1, 3]
There are relative minima at about
(–0.61, –0.18) and (0.6, –0.18); there are no
relative maxima (The function is not defined
at x = 0.) There are inflection points at about
(–0.22, –0.07) and (0.22, –0.07)
on [–2, 2] by [–2, 2]
There is a relative maximum at about
(–0.37, 0.37); relative minimum at about
(0.37, –0.37). There are no inflection points.
(The function is not defined at x = 0.)
67.
y 2 − ye x = 12
d
dx
( y 2 − ye x ) = dxd (12)
68.
d
dx
2 yy '− ( y ' e x + ye x ) = 0
y '(2 y − e x ) = ye x
y' =
69.
y'
y '(2 y − xy ) = ln y
2 y −e x
f '( x ) = 40,000(0.195)e 0.195 x
f '( x ) = 7800e 0.195 x
f '(1) = 7800e 0.195(1)
f '(1) = 9479.425695
Annual salary increases by about $9480 per
extra year of calculus.
© 2010 Brooks/Cole, Cengage Learning.
( y 2 − x ln y ) = dxd (10)
2 yy '− (ln y + x y ) = 0
ye x
f ( x ) = 40,000e 0.195 x
y 2 − x ln y = 10
y' =
70.
ln y
2 y − xy
=
y ln y
2 y2 − x
2015 – 2005 = 10 = t
f (t ) = 5e 0.24t
f '(t ) = 5(0.24)e 0.24t
f '(t ) = 1.2e 0.24t
f '(10) = 1.2e 0.24(10)
f '(10) = 13.22781166
The number of thousand megawatts increases
by about 13.2 per extra year.
Exercises 4.3
189
71.
a.
To find the rate of growth after 0 years,
evaluate V ′ (0) .
V (t ) = 1000e0.05t
V ′(t ) = 1000e0.05t (0.05) = 50e0.05t
V ′(0) = 50e0.05(0) = 50e0 = 50
The rate of growth after 0 years is $50 per
year.
b. To find the rate of growth after 10 years,
evaluate V ′ (10) .
V ′(t ) = 50e0.05t
V ′(10) = 50e0.05(10) = 50e0.5 ≈ 82.44
The rate of growth after 10 years is $82.44
per year.
72.
a.
73.
P (t ) = 6.45e0.0175t
P ′(t ) = 6.45e0.0175t (0.0175) = 0.112875e0.0175t
In 2015, t = 2015 – 2005 = 10.
74.
P (t ) = 75e −0.2t + 25
P ′(t ) = 75e −0.2t (−0.2) = −15e −0.2t
0.0175(10)
P ′(10) = 0.112875e
≈ 0.134
In 2015, the world population is growing by 134
million people per year.
75.
A(t ) = 1.2e −0.05t
A′(t ) = 1.2e −0.05t (−0.05) = −0.06e −0.05t
A′(0) = −0.06e −0.05(0) = −0.06e0 = −0.06
The amount remaining after 0 hours is
decreasing by 0.06 mg per hour.
b. A′(2) = −0.06e −0.05(2) = −0.06e −0.1
≈ −0.054
The amount remaining after 2 hours is
decreasing by 0.054 mg per hour.
a.
77.
76.
a.
P ′(0) = −15e −0.2(0) = −15
It is decreasing by 15% per time unit.
b.
P′ (3) = −15e −0.2(3) ≈ −8.2
It is decreasing by 8.2% per time unit.
T (t ) = 70 + 130e −2.5t
T ′(t ) = −325e −2.5t
a. T ′(0) = −325
The temperature is decreasing by 325
degrees per hour.
b. T ′(1) = −325e −2.5(1) ≈ −27
The temperature is decreasing by 27 degrees
per hour.
S ( x) = 1000 − 900e−0.1x
← weekly sales (in thousands)
S ′( x) = −900e−0.1x (−0.1) ← rate of change of sales per week
= 90e−0.1x
a. S ′(1) = 90e−0.1
← rate of change of sales per week after 1 week
≈ 90(0.9048)
≈ 81.4 thousand sales per week
b.
78.
To find the rate of change at t = 0 evaluate
V ′ (0) .
V (t ) = 10, 000e−0.35t
V ′(t ) = 10, 000e−0.35t (−0.35)
= −3500e−0.35t
V ′(0) = −3500e−0.35(0) = −3500
The value is decreasing by $3500 per year.
b. To find the rate of change after 2 years,
evaluate V ′ (2) .
V ′(2) = −3500e −0.35(2) ≈ −1738.05
The value is decreasing by $1738.05 per
year.
S ′(10) = 90e−1 ≈ 90(0.3679) ≈ 33 thousand sales per week after 10 weeks
N (t ) = 50, 000(1 − e−0.4t )
N ′(t ) = −50, 000e−0.4t (−0.4) = 20, 000e−0.4t
a.
N ′(0) = 20, 000e−0.4(0) = 20, 000 people per hour
b.
N ′(8) = 20, 000e−0.4(8) ≈ 815 people per hour
© 2010 Brooks/Cole, Cengage Learning.
190
79.
Chapter 4: Exponential and Logarithmic Functions
To find the maximum consumer expenditure, solve E′ ( p) = 0 , where E ( p ) = pD( p ) .
E ( p) = pD( p ) = 5000 pe −0.01 p
E ′( p ) = 5000e −0.01 p + 5000 pe −0.01 p (−0.01)
= 5000e −0.01 p − 50 pe −0.01 p
Setting E′ ( p) = 0 , we get
5000e −0.01 p − 50 pe −0.01 p = 0
50e −0.01 p (100 − p) = 0
p = $100
We use the second derivative test to show that p = 100 is a maximum.
E ′′( p ) = 5000e −0.01 p (−0.01) − 50e −0.01 p − 50 pe −0.01 p (−0.01)
= −50e −0.01 p − 50e −0.01 p + 0.5 pe −0.01 p
= −100e −0.01 p + 0.5 pe −0.01 p
E ′′(100) = −100e −1 + 50e −1 = −50e −1 < 0
so E( p) is maximized.
80.
To find the maximum consumer expenditure, solve E′ ( p) = 0 , where E( p) = pD(p) .
E ( p ) = pD( p) = 8000 pe −0.05 p
E ′( p ) = 8000e −0.05 p + 8000 pe −0.05 p (−0.05)
0 = 8000e −0.05 p − 400 pe −0.05 p
0 = (20 − p)e −0.05 p
p = $20
Use the second derivative test to show that p = 20 is a maximum.
E ′′( p ) = 8000e−0.05 p (−0.05) − 400e −0.05 p − 400 pe −0.05 p (−0.05)
= −400e −0.05 p − 400e−0.05 p + 20 pe−0.05 p
= −800e −0.01 p + 20 pe −0.01 p
E ′′(20) = −800e −0.2 + 400e −0.2 = −400e −0.2 < 0
so E( p) is maximized.
81.
p( x ) = 400e −0.20 x
← price function (in dollars)
← revenue function
a. R ( x ) = xp( x )
= 400 xe −0.20 x
b. To maximize R(x) , differentiate.
R ′( x ) = 400e −0.20 x + 400 xe −0.20 x ( −0.20) = 400e −0.20 x (1 − 0.2 x )
R ′( x ) = 0 when x = 1 = 5, which is the only critical value.
0.2
We calculate R′′ (x) for the second derivative test.
R ′′( x ) = 400e −0.20 x ( −0.20)(1 − 0.2 x ) + 400e −0.20 x ( −0.20)
= 400e −0.20 x ( −0.20)(1 − 0.2 x + 1)
= −80e −0.20 x (2 − 0.2 x )
= −160e −0.20 x (1 − 0.1x )
At the critical value of x = 5 ,
R ′′(5) = −80e −1 < 0 ,
so R(x) is maximized at x = 5..
At x = 5, p( x ) = 400e −0.20 x
p(5) = 400e −1 ≈ 147.15
The revenue is maximized at quantity x = 5000 and price p = $147.15.
© 2010 Brooks/Cole, Cengage Learning.
Exercises 4.3
82.
191
a. R ( x ) = xp ( x ) = 4 x − x ln x
b. To maximize, differentiate and solve R′ (x) = 0 .
R ′( x ) = 4 − ln x − x 1x = 3 − ln x
( )
0 = 3 − ln x
ln x = 3
x = e3 ≈ 20.086
Now use the second derivative test.
R ′′( x ) = − 1x < 0 for x > 0, so R( x ) is maximized.
At x = e3 , p ( x ) = 4 − ln e3 = 1
The revenue is maximized at
x = 20,086 units and price $1.
83. a.
T ( x ) = 70 − 30e −3.5t
T (0.25) = 70 − 30e −3.5(0.25) ≈ 57.5
T ′( x ) = −30e −3.5t ( −3.5) = 105e −3.5t
T ′(0.25) = 105e −3.5(0.25) ≈ 43.8
After 15 minutes, the temperature of the beer
is 57.5 degrees and is increasing at the rate
of 43.8 degrees per hour.
b. T ( x ) = 70 − 30e −3.5t
T (1) = 70 − 30e −3.5(1) ≈ 69.1
T ′( x ) = −30e −3.5t ( −3.5) = 105e −3.5t
T ′(1) = 105e −3.5(1) ≈ 3.2
After 1 hour, the temperature of the beer is
69.1 degrees and is increasing at the rate of
3.2 degrees per hour.
85.
A speed of 10 meters per second after 2.31
seconds.
84. a.
b.
N ( x ) = 200, 000 (1 − e −0.5t )
N (0.5) = 200, 000 (1 − e −0.5(0.5) ) ≈ 44, 240
N ′( x ) = 200, 000 ( − e −0.5t ) ( −0.5) = 100, 000e −0.5t
N ′(0.5) = 100,000e −0.5(0.5) ≈ 77,880
After 30 minutes, the number of people who have
heard the bulletin is about 44,240 and is increasing
at the rate of 77,880 per hour.
N ( x ) = 200,000 (1 − e −0.5t )
N (3) = 200,000 (1 − e −0.5(3) ) ≈ 155,374
N ′( x ) = 200,000 ( − e −0.5t ) ( −0.5) = 100,000e −0.5t
N ′(3) = 100, 000e −0.5(3) ≈ 22, 313
After 3 hours, the number of people who have
heard the bulletin is about 155,374 and is
increasing at the rate of 22,313 per hour.
86.
on [0.9.93] by [0,20]
After 0.1 seconds, Lewis’s acceleration was 9.678
meters per second2. After 9.33 seconds, his
acceleration was 0.002 meters per second2.
87
a.
88. The function e x cannot be differentiated by the
power rule because he base e, is a constant and
the variable x, is in the exponent.
on [35,
[35, 80]
240]
on
80]byby[0,[0,
240]
f (35) ≈ 132.4; f ′(35) ≈ 0.834
At age 35, the fastest man’s time is 2
hours 12.4 minutes and increasing at
about 0.834 minutes per year.
c. f (80) ≈ 234.9; f ′(80) ≈ 7.76
At age 80, the fastest man’s time is 3
hours 54.9 minutes and increasing at
about 7.76 minutes per year.
b.
© 2010 Brooks/Cole, Cengage Learning.
192
Chapter 4: Exponential and Logarithmic Functions
89.
b. d e x = e x
dx
90.
a. d e f ( x ) = f '( x ) ⋅ e f ( x )
dx
91.
c. d e5 = d 5 ⋅ e5
dx
dx
d e 5 = 0 ⋅ e5
dx
d e5 = 0
dx
92.
a. d x e = ex e −1
dx
93.
b. d ln x = 1
dx
x
94.
f '( x )
b. d ln f ( x ) =
dx
f ( x)
95.
d 5
d
dx
c.
ln 5 =
dx
5
d ln 5 = 0
dx
5
d ln 5 = 0
dx
96.
f '( x )
False: d ln f ( x ) =
. It does not involve a
dx
f ( x)
natural log.
97.
e ln x = x and d x = 1
dx
98.
ln e x = x and d x = 1
dx
99.
N (t ) = Ke− ae
− bt
= Ke− ae
− bt
= Ke− ae
− bt
dN
dt
= bN (ae
for t ≥ 0, k > 0, b > 0.
d
dt
(
−bt
( −ae−bt )
− ae−bt (−b)
100.
y = axe − bx for x ≥ 0, a > 1, b > 0.
y ′ = a ( e − bx + xe − bx ( −b))
= ae − bx (1 − bx )
y ′ = 0 when 1 − bx = 0.
y ′ = 0 at x = 1b .
Second derivative test
y ′′ = a ( −b)e − bx (1 − bx ) + ae − bx ( −b)
= abe − bx ( −1 + bx − 1) = abe − bx (bx − 2)
)
)
ae− bt
= bN (ln e
)
⎛ K ⎞
= bN ln ⎜
− bt ⎟
⎝ Ke− ae ⎠
= bN ln K
N
( )
( NK ) < 0 and the population will be
decreasing. If N < K, ln ( NK ) > 0 and the
If N > K, ln
population will be increasing.
© 2010 Brooks/Cole, Cengage Learning.
y ′′
( 1b ) = abe−b(
=
− ab
e
1
b
<0
) (b
( 1b ) − 2) = abe−1 (1 − 2)
( a > 1, b > 0)
This shows that 1 is indeed a maximum.
b
The population is maximized when the
parental stock is 1 .
b
Exercises 4.3
193
101. To maximize R(r ) , solve R′ (r ) = 0.
R ′( r ) = ar − b = 0
a − br = 0
r = ab
102.
To find the maximum concentration, solve
A′ (t ) = 0 , where
A(t ) =
=
Using the second derivative test,
R ′′( r ) = − a2 < 0
0.1 ( e −0.4 t
0.6−0.4
e −0.4 t − e −0.6 t
2
A′(t ) =
− e −0.6t )
e −0.4 t ( −0.4) − e −0.6 t ( −0.6)
2
0 = 0.3e −0.6t − 0.2e −0.4 t
0.2e −0.4t = 0.3e −0.6t
ln(0.2e −0.4t ) = ln(0.3e −0.6t )
ln(0.2) + ( −0.4t ) = ln 0.3 + ( −0.6t )
0.2t = ln 0.3 − ln 0.2
−ln 0.2 ≈ 2.03
t = ln 0.30.2
r
so R(r ) is maximized.
Now use the second derivative test.
A′′(t ) = 0.3e −0.6t ( −0.6) − 0.2e −0.4 t ( −0.4)
= 0.08e −0.4 t − 0.18e −0.6 t
A′′(2.03) = 0.08e −0.4(2.03) − 0.18e −0.6(2.03)
≈ −0.0178
so A(t ) is maximized.
The maximum concentration occurs at about
2.03 hours.
103 a.
y = log a x
x = ay
ln x = ln a y
ln x = y ln a
ln x = y
ln a
ln x = log x
a
ln a
for a > 0 and x > 0
b.
y = ln x
ln a
y ln a = ln x
ln a y = ln x
ay = x
log a x = y
log a = ln x
ln a
104.
a x = ( eln a ) since x = eln x
= e(ln a ) x
105.
2.23x = ( eln 2.23 ) = e(ln 2.23) x ≈ e0.802 x
106.
0.63x = ( eln 0.63 ) = e(ln 0.63) x ≈ e −0.462 x
107.
a.
x
x
x
b.
c.
d.
e.
d 10 x = (ln 10)10 x
dx
d 3x 2 +1 = (ln 3)3x 2 +1 (2 x ) = 2(ln 3) x 3x 2 +1
dx
d 23 x = (ln 2)23 x (3) = (3ln 2)23 x
dx
d 53 x 2 = (ln 5)53 x 2 (6 x ) = 6(ln 5) x 53 x 2
dx
d 24− x = (ln 2)24− x ( −1) = ( − ln 2) 24− x
dx
© 2010 Brooks/Cole, Cengage Learning.
194
Chapter 4: Exponential and Logarithmic Functions
108. a.
b.
c.
d.
e.
d 5 x = (ln 5)5 x
dx
d 2 x 2 −1 = (ln 2)2 x 2 −1 (2 x ) = 2(ln 2) x 2 x 2 −1
dx
d 34 x = (ln 3)34 x (4) = (4 ln 3)34 x
dx
d 95 x 2 = (ln 9)95 x 2 (10 x ) = 10(ln 9) x 95 x 2
dx
d 101− x = (ln10)101− x ( −1) = ( − ln10)101− x
dx
110. a. log x =
3
109.
d log x = 1
2
dx
(ln 2) x
2
d
2x
b.
log ( x − 1) =
dx 10
(ln 10)( x 2 − 1)
a.
c.
d log ( x 4 − 2 x ) =
4 x3 − 2
3
dx
(ln 3)( x 4 − 2 x )
1
(ln 3) x
b. log 2 ( x 2 + 1) =
2x
(ln 2)( x 2 + 1)
c. log10 ( x 3 − 4 x ) =
3x 2 − 4
(ln 10)( x 3 − 4 x )
EXERCISES 4.4
1.
3.
a.
ln f (t ) = ln t 2 = 2 ln t
d ln f (t ) = d 2 ln t =
dt
dt
2.
b.
For t = 1, the relative rate of change is
2 =2.
1
For t = 10, the relative rate of change is
2 = 0.2 .
10
a.
ln f (t ) = ln100e0.2 t
= ln100 + ln e0.2 t
= ln100 + 0.2t
d ln f (t ) = 0.2
dt
b.
4.
a.
2
ln f (t ) = ln et = t 2
d ln f (t ) = 2t
dt
6.
b. For t = 10, the relative rate of change is
2(10) = 20.
7.
a.
2
ln f (t ) = ln e − t = −t 2
d ln f (t ) = −2t
dt
b. For t = 10 , the relative rate of change is
–2(10) = –20.
© 2010 Brooks/Cole, Cengage Learning.
a.
ln f (t ) = ln t 3 = 3ln t
d ln f (t ) = d 3ln t =
dt
dt
3
t
For t = 1, the relative rate of change is
3 = 3.
1
For t = 10, the relative rate of change is
3 = 0.3 .
10
ln f (t ) = ln100e −0.5t
= ln100 + ln e −0.5t
= ln100 − 0.5t
d ln f (t ) = −0.5
dt
b. For t = 4 , the relative rate of change is
–0.5.
b. For t = 5, the relative rate of change is 0.2.
5.
a.
2
t
a.
3
ln f (t ) = ln et = t 3
d ln f (t ) = 3t 2
dt
b. For t = 5, the relative rate of change is
3(5)2 = 75 .
8.
a.
3
ln f (t ) = ln e − t = −t 3
d ln f (t ) = −3t 2
dt
b. For t = 5, the relative rate of change is
−3(5)2 = −75 .
Exercises 4.4
9.
11.
ln f (t ) = ln 25 t − 1
= ln 25(t − 1)1/ 2
= ln 25 + 1 ln(t − 1)
2
d ln f (t ) = 0 + 1
dt
2(t − 1)
b. For t = 6 , the relative rate of change is
1
1 .
= 10
2(6−1)
10.
A( x ) = 3.2e0.5 x
ln A( x ) = ln(3.2e0.5 x ) = ln 3.2 + ln e0.5 x
= ln 3.2 + 0.5 x
d ln A( x ) = 0.5
dx
12.
a.
ln f (t ) = ln100 3 t + 2
= ln100 + 1 ln(t + 2)
a.
3
d ln f (t ) = 0 + 1
3(t + 2)
dt
b. For t = 8 , the relative rate of change is
1 = 1 .
3(10)
30
B ( x ) = 32e0.11x
ln B ( x ) = ln(32e0.11 x ) = ln 32 + ln e0.11x
= ln 32 + 0.11x
d ln B ( x ) = 0.1
dx
A′( x ) 3.2e0.5 x (0.5)
=
= 0.5
A( x )
3.2e0.5 x
For t = 10, the relative rate of change is
d ln A(10) = 0.5 or 50% . The stock is
dx
B ′( x ) 32e0.11x (0.11)
=
= 0.11
B( x )
32e0.11x
For t = 11, the relative rate of change is
d ln B (11) = 0.11 or 11% . The stock is
dx
appreciating by 50% per year.
appreciating by 11% per year.
d ln
dx
d
dt
d ln B ( x )
dx
A( x ) =
ln N (t ) = ln(0.5 + 1.1e0.01t )
13.
ln N (t ) =
ln N (t ) = ln(0.4 + 1.2e0.01t )
14.
1.1e0.01t (0.01)
d
dt
0.5+1.1e0.01t
=
ln N (t ) =
1.2 e0.01t (0.01)
0.4 +1.2 e0.01t
For t = 10, the relative rate of change is
For t = 10, the relative rate of change is
1.1e0.01(10) (0.01)
1.2 e0.1 (0.01)
0.5+1.1e
15.
195
0.01(10)
≈ 0.0071 or 0.71% .
ln P (t ) = ln(4 + 1.3e0.04t )
a.
d
dt
ln P (t ) =
1.3e0.04 t (0.04)
4+1.3e
0.04 t
=
0.052 e0.04 t
4 +1.3e0.04 t
For t = 8, the relative rate of change is
0.04(8)
0.052e
≈ 0.012 or 1.2% .
4 + 1.3e0.04(8)
b. If the relative rate of change is
1.5% = 0.0015, then
0.052 e0.04 t
4 +1.3e0.04 t
0.4 +1.2 e0.1
16.
≈ 0.0077 or 0.77% .
ln P (t ) = ln(6 + 1.7e0.05t )
a.
d
dt
ln P (t ) =
1.7 e0.05t (0.05)
6+1.7 e0.05t
=
0.085e0.05t
6+1.7 e0.05t
For t = 8, the relative rate of change is
0.085e0.05(8) ≈ 0.015 or 1.5% .
6+1.7e0.05(8)
b. The relative rate of change will reach 1.5%
in about 8.3 years.
= 0.015
0.052e
= 0.06 + 0.0195e0.04 t
0.04 t
0.0325e
= 0.06
0.04 t
0.06
= 0.0325
e
0.04 t
0.06
( 0.0325
)
0.06 ⎤ ≈ 15.3
1
t = 0.04 ⎡⎣ ln ( 0.0325
)⎦
0.04t = ln
The relative rate of change will reach 1.5%
in about 15.3 years.
© 2010 Brooks/Cole, Cengage Learning.
196
17.
Chapter 4: Exponential and Logarithmic Functions
D ( p ) = 200 − 5 p, p = 10
18.
Elasticity of demand is
5p
− pD ′( p ) − p( −5)
E ( p) =
=
=
200 − 5 p 200 − 5 p
D( p)
p
=
40 − p
b. Evaluating at p = 10 ,
a.
− p ( −8)
60−8 p
=
8p
60−8 p
2p
a.
E ( p) =
= 15−2 p
b.
E (5) = 15−2(5) = 10
=2
5
2(5)
Since E(5) > 1 , the demand is elastic.
10 = 10 = 1
40 − 10 30 3
The elasticity is less than 1, and so the
demand is inelastic at p = 10 .
E (10) =
19.
− p ( −2 p )
a.
E ( p) =
b.
E (10) =
300− p
2 p2
=
2
2(10)2
300− p
=
300−(10)2
200
200
20.
2
=1
a.
E ( p) =
b.
E (5) =
Since E (10) = 1 , the demand is unitary
elastic.
21.
a.
E ( p) =
− p ( − 300
)
2
p
300
p
=
300
p
300
p
E ( p) =
=1
22.
=
b.
− p[ 12 (175−3 p ) −1/ 2 ]( −3)
E (50) =
3(50)
2[175−3(50)]
= 150
=3
50
25.
a.
E ( p) =
=
200
p2
100
p2
b.
=2
26.
− p (4000e −0.01 p )( −0.01)
a.
E ( p) =
b.
E (200) = 0.01(200) = 2
Since E (200) > 1 , the demand is elastic.
4000e
−0.01 p
= 0.01 p
© 2010 Brooks/Cole, Cengage Learning.
2 p2
100− p 2
= 50
=
75
2
3
E ( p) =
− p ( − 500
)
2
p
500
p
=
500
p
500
p
=1
E ( p) =
− p[ 12 (100 − 2 p ) −1/ 2 ]( −2)
(100 − 2 p )1/ 2
E (20) = 10020−40 =
20
60
=
p
100 − 2 p
= 13
Since E (20) < 1 , the demand is inelastic.
b. Since E (40) > 1 , the demand is elastic.
27.
2(5)2
100−(5)2
=
b. Since E (2) = 1 , the demand is unitary
elastic.
Since E (50) > 1 , the demand is elastic.
⎛
⎞
− p⎜⎜ − 200
3 ⎟⎟
p
⎝
⎠
100
p2
a.
24. a.
(175−3 p )1/ 2
3p
2(175−3 p )
100− p
2
Since E (5) < 1 , the demand is inelastic.
b. Since E (4) = 1 , the demand is unitary elastic.
23. a.
− p ( −2 p )
a.
E ( p) =
⎛
⎞
− p ⎜⎜ −1800
4 ⎟⎟
p
⎝
⎠
600
p3
=3
b. Since E (25) > 1 , the demand is elastic.
28.
− p (6000e −0.05 p )( −0.05)
a.
E ( p) =
b.
E (100) = 5
Since E (100) > 1 , the demand is elastic.
6000e −0.05 p
= 0.05 p
Exercises 4.4
29.
31.
33.
197
The demand function is D ( p ) = 2(15 − 0.001 p )2
To determine whether the dealer needs to raise or
lower the price to increase revenue, we need to
determine the elasticity of demand.
− pD ′( p )
E ( p) =
D( p )
− p ⋅ 4(15 − 0.001 p )( −0.001)
=
2(15 − 0.001 p )2
0.002 p
=
15 − 0.001 p
When the cars sell at a price of $12,000,
0.002(12,000)
E (12,000) =
= 24 = 24 = 8
15 − 0.001(12, 000) 15 − 12 3
Since E = 8 > 1 , to increase revenues the dealer
should lower prices.
30.
To determine the elasticity of demand, we
consider
− p{ 12 [150,000(1.75 − p ) −1/ 2 ( −1)]}
E ( p) =
150, 000(1.75 − p )1/ 2
p
=
2(1.75 − p )
When the fare is 75 cents,
1.25
E (1.25) =
2(1.75 − 1.25)
= 1.25 = 1.25
1
Since demand is elastic, raising the fare will
not succeed.
32.
D ( p ) = 120
10 + p
− pD ′( p )
=
E ( p) =
D( p)
To determine the elasticity of demand, we
consider
E ( p) =
When the price is $15,
E (15) = 15 = 3
5
Since demand is elastic, the discounts will
succeed.
To determine the elasticity of demand, we
consider
E ( p) =
=
−120
(10+ p )2
120
10+ p
=
p
10 + p
− p(3.5e −0.06 p )( −0.06)
= 0.06 p
3.5e −0.06 p
When the price is $120 per barrel,
E (120) = 0.06(120) = 7.2
Since demand is elastic, it should lower prices.
E ( p) =
To determine the elasticity of demand, we
consider
E ( p) =
6 = 6 = 3 = 0.375
10 + 6 16 8
Since E = 3 < 1 , increasing prices should increase
8
revenues. Yes, the commission should grant the
request.
To determine the elasticity of demand, we
consider
80,000(75− p )1/ 2
p
2(75− p )
Since E(50) = 1, raising prices will not
succeed.
− p(9.5e −0.04 p )( −0.04)
= 0.04 p
9.5e −0.04 p
When the price is $120 per barrel,
E (120) = 0.04(120) = 4.8
Since demand is elastic, it should lower prices.
E (6) =
35.
− p[40,000(75− p ) −1/ 2 ( −1)]
When the price is 50 cents,
E (50) = 2(7550−50) = 1
34.
−p⋅
3p
p
− p( −3)
=
=
60 − 3 p 60 − 3 p 20 − p
36.
E ( p) =
=
− p (3.509)( −0.859) p −1.859
3.509 p −0.859
0.859 p
= 0.859
p
© 2010 Brooks/Cole, Cengage Learning.
198
37.
39.
Chapter 4: Exponential and Logarithmic Functions
− p(7.881)( −0.112 p −1.112 )
7.881 p −0.112
0.112 p
=
p
= 0.112
38.
E ( p) =
a.
E = 0.75
b. Since demand is inelastic, raising prices will
raise revenue.
c. $13,333
a. E = 0.35
b. Since demand is inelastic, raising prices will
raise revenue.
c. $20,401
40.
a.
b.
E ( p) =
− p ( −1)
90− p
E (0) =
0
90− 0
c.
E (89.999) =
=
p
90− p
=0
89.999 ≈ 89,999
90−89.999
≈ 89,999, 999
E (89.999999) = 9089.999999
−89.999999
As p approaches 90, E approaches infinity.
d. The midpoint is p = 45.
E (45) = 9045
=1
− 45
e. For D ( p ) = a − bp ,
E ( p) =
− p( − b)
a − bp
=
bp
a −bp
(b) For p = 0, E (0) =
b(0)
a − b(0)
=0.
(c) When the value p approaches b, E
approaches infinity.
(d) For the midpoint, p =
E
41.
E=
percent change in demand
percent change in price
52 − 47.5
52
=
≈ 0.08654 = 8.65%
2.80 − 2.40 0.1667 16.7%
2.40
≈ 0.52 or 52%
42.
43.
f ( x ) = e 1.2 x
d ln f ( x ) = d ln e 1.2 x = d 1.2 x = 1.2
dx
dx
dx
44.
45.
f ( x) = x n
d x
d ln f ( x ) = d ln x n = dx
dx
dx
xn
46.
n
n −1
= nx
xn
=n
x
© 2010 Brooks/Cole, Cengage Learning.
a
b = a
2 2b
,
( ) a a
( ) = ( ) = a−2a = 2a2−a = aa = 1
2
2
b a
2b
a −b a
2b
a
2b
f ( x) = e x
d ln f ( x ) = d ln e x = d x = 1
dx
dx
dx
f ( x) = x
d ln f ( x ) = d ln x = d ln x = 1
dx
dx
dx
x
(
)
−p d e−p
dx
E ( p) =
e −2 p
− p( −1)( e − p )
E ( p) =
e−p
E ( p) = p
Review Exercises for Chapter 4
(
)
199
47.
− p d e −2 p
dx
E ( p) =
e −2 p
− p( −2)( e −2 p )
E ( p) =
e −2 p
E ( p) = 2 p
48.
Demand will increase weekly.
49.
Demand will increase strongly.
50.
Elastic for heating oil and inelastic for olive oil.
(It is commonly used.)
51.
Inelastic for cigarettes (they are habit-forming)
and elastic for jewelry.
52.
a.
⎛
⎞ nc
− p ⎜⎜ −nnc
+1 ⎟⎟
p ⎠ pn
E ( p) = ⎝
=
=n
c
c
n
n
p
p
b. If E ( p ) = 1 , then we know from part (a) that
D ( p ) = c1 = c .
p
p
53.
E ( p) =
− p(ae − cp )( − c)
= cp
ae − cp
54.
If S ( p ) = aecp , then
Es ( p ) =
55.
pS ′( p ) p( aecp )( c)
=
= cp
S ( p)
aecp
If S ( p ) = ap n , then
Es ( p ) =
pS ′( p ) p( anp n −1 ) np n
=
= n =n
S ( p)
ap n
p
REVIEW EXERCISES FOR CHAPTER 4
1.
a.
For quarterly compounding, r = 0.08 = 0.02
4
and n = 8 ⋅ 4 = 32 .
2.
P(1 + r )n = 10,000(1 + 0.02)32
= 10,000(1.02)32 ≈ $18,845.41
b.
P (1 + r )n = 1(1 + 0.015)4 ≈ 1.0614
For 5.98% compounded continuously, r = 0.0598
and n = 1.
For compounding continuously, r = 0.08 and
n = 8 ⋅ 4 = 32 .
Pe rn = e0.0598 ≈ 1.0616
5.98% compounded continuously has a higher
yield.
Pe rn = 10, 000e(0.08)(8) ≈ $18,964.81
3.
a.
If the depreciation is 20% per year,
r = −20% = −0.2 and n = t . The formula
for the value is
V = 800,000(1 − 0.2)t = 800, 000(0.8)t
b. After 4 years, its value is
V = 800, 000(0.8)4 = $327,680
For 6% compounded quarterly, r = 0.06 = 0.015
4
and n = 4 · 1 = 4.
4.
For Drug A, C (t ) = 2e −0.2t . After 4 hours,
C (4) = 2e −0.8 ≈ 0.899
For Drug B, C (t ) = 3e −0.25t . After 4 hours,
C (4) = 3e −1 ≈ 1.103
Drug B has a higher concentration.
© 2010 Brooks/Cole, Cengage Learning.
200
Chapter 4: Exponential and Logarithmic Functions
5.
6.
The value t = 2011 – 1987 = 24 corresponds to
2011 and C (24) = 424 / 3 = 48 = 65,536 megabits .
São Paulo will overtake Tokyo in about 60 years,
or during the year 2060.
7.
9.
r = 12 10% = 5% = 0.05
a.
P (1 + 0.05)n = 2 P
1.05n = 2
ln1.05n = ln 2
n ln1.05 = ln 2
n = ln 2 ≈ 0.6931 ≈ 14.2
ln1.05 0.0488
Since n is in half years, we divide by 2 to
convert to years. About 7.1 years.
b.
P (1.05)n = 1.5P
1.05n = 1.5
ln1.05n = ln1.5
n ln1.05 = ln1.5
n = ln1.5
ln1.05
≈ 0.4055
0.0488
≈ 8.31 half years
8.31 ≈ 4.2 years
2
The proportion of potassium 40 remaining after t
8.
b. To increase by 50%,
Pe0.07 n = 1.5P
e0.07 n = 1.5
n = ln1.5 ≈ 5.8 years
0.07
10.
© 2010 Brooks/Cole, Cengage Learning.
Since 99.9% = 0.999 remains,
e −0.00054t = 0.999
t = ln 0.999 ≈ 1.85 million years old
−0.00054
e −0.00054t = 0.973
−0.00054t = ln 0.973
t = ln 0.973 ≈ 50.7 million years old
−0.00054
We want to solve N (t ) = 1,000, 000(1 − e −0.3t )
with N(t) = 500,000 .
1,000,000(1 − e −0.3t ) = 500,000
1 − e −0.3t = 0.5
−e −0.3t = −0.5
e −0.3t = 0.5
−0.3t = ln 0.5
0.5 ≈ −0.6931
t = ln−0.3
−0.3
≈ 2.3 hours
For compounding continuously,
r = 7% = 0.07.
Pe0.07 n = 2 P
e0.07 n = 2
0.07n = ln 2
n = ln 2 ≈ 9.9 years
0.07
million years is e −0.00054t .
Since 97.3% = 0.973 remains,
11.
a.
12.
Since the rate of increase is 3% = 0.03 per year,
r = 0.03. To find when demand will increase by
50% = 0.5,
1.5P = P (1 + 0.03)n
1.5 = (1.03)n
ln 1.5
n=
≈ 13.7 years
ln 1.03
Review Exercises for Chapter 4
13.
a.
201
If the interest rate is 6.5% = 0.065
compounded quarterly, r = 0.065 .
4
(
(1 +
1000 1 +
)
)
0.065 n
4
0.065 n
4
14.
a.
0.3 = 1 − e −0.032t
e
= 0.7
−0.032t = ln 0.7
0.7 ≈ 11 days
t = −ln0.032
= 1.5(1000)
−0.032t
= 1.5
n=
ln1.5
ln 1+ 0.065
4
(
To reach 30% = 0.3 of the people,
p(t) = 0.3 .
)
≈ 25.2 quarters
≈ 6.3 years
b. If the interest rate is 6.5% = 0.065
compounded continuously, r = 0.065.
1000e0.065n = 1.5(1000)
e0.065n = 1.5
0.065n = ln1.5
ln1.5 ≈ 6.24 years
n = 0.065
b. To reach 40% = 0.4 of the people,
p(t ) = 0.4 .
0.4 = 1 − e −0.032t
e −0.032t = 0.6
0.6 ≈ 16 days
t = −ln0.032
2
d ln( x 2 − 1)2 = 2( x − 1)(2 x )
dx
( x 2 − 1)2
= 24 x
x −1
15.
d ln 2 x = 2 = 1
dx
2x x
16.
17.
d ln(1 − x ) = − 1
dx
1− x
18.
19.
1 −2 / 3
d ln 3 x = d ln x1/ 3 = 3 x
dx
dx
x1/ 3
1
=
3x
20.
d ln e x = d ( x ) = 1
dx
dx
21.
d ln x 2 = d (2 ln x ) = 2
dx
dx
x
22.
d ( x ln x − x ) = ln x + x 1 − 1
dx
x
= ln x
23.
d e − x 2 = e − x 2 ( −2 x ) = −2 xe − x 2
dx
24.
d e1− x = e1− x ( −1) = − e1− x
dx
25.
d ln e x 2 = d ( x 2 ) = 2 x
dx
dx
26.
d (e x 2 ln x − x 2 / 2 ) = e x 2 ln x − x 2 / 2 ⎡ 2 x ln x + x 2 1 − x ⎤
⎥⎦
dx
x
⎣⎢
28.
d (2 x 3
dx
1 ( x 2 + 1) −1/ 2 (2 x )
d ln x 2 + 1 = 2
dx
( x 2 + 1)1/ 2
= 2x
x +1
()
()
2
2
= 2 x ln x e x ln x − x / 2
27.
d (5 x 2
dx
+ 2 x ln x + 1) = 10 x + 2 ln x + 2 x
= 10 x + 2 ln x + 2
( 1x )
()
+ 3 x ln x − 1) = 6 x 2 + 3ln x + 3x 1x
= 6 x + 3ln x + 3
2
© 2010 Brooks/Cole, Cengage Learning.
202
29.
Chapter 4: Exponential and Logarithmic Functions
d (2 x 3
dx
− 3 xe 2 x ) = 6 x 2 − 3e 2 x − 3 xe2 x (2)
30.
= 6 x 2 − 3e 2 x − 6 xe 2 x
31.
= 4 − 4 xe2 x − 2 x 2 e 2 x (2)
= 4 − 4 xe 2 x − 4 x 2 e 2 x
32.
Relative minimum at (0, 1.4)
33.
d (4 x − 2 x 2 e 2 x )
dx
a.
S ( x ) = 2000 − 1500e −0.1x
S ′( x ) = −1500e −0.1x (−0.1) = 150e −0.1x
S ′(1) = 150e −0.1(1) ≈ 136
Sales are increasing by 136,000 after
1 week.
b.
S ′(10) = 150e −0.1(10) = 150e −1 ≈ 55
Sales are increasing by 55,000 after 10
weeks.
Relative maximum at (0, 16).
34.
a.
A(t ) = 1.5e −0.08t
A′(t ) = 1.5e −0.08t (−0.08) = −0.12e −0.08t
A′(0) = −0.12e −0.08(0) = −0.12
The amount of the drug in the
bloodstream immediately after the
injection is decreasing by 0.12 mg per
hour.
b. A′(5) = −0.12e −0.08(5) ≈ −0.08
The amount of the drug in the
bloodstream after 5 hours is decreasing by
0.08 mg per hour.
35.
P(t ) = 100 − 200 ln(t + 1)
← rate of change
P ′(t ) = −200
t +1
P ′(5) = −200 = −200 ← rate of change after 5 seconds
5 +1
6
= −100 = −33 1
3
3
The rate of change after 5 seconds is decreasing by 33 1 % per second.
3
36.
a.
T (t ) = 70 − 35e −0.1t
T ′( x ) = −35e −0.1t ( −0.1) = 3.5e −0.1t
T ′(0) = 3.5e0 = 3.5 degrees per hour
b.
T ′(5) = 3.5e −0.5 ≈ 2.1 degrees per hour
© 2010 Brooks/Cole, Cengage Learning.
37.
N (t ) = 30, 000(1 − e −0.3t )
N ′(t ) = −30, 000e −0.3t ( −0.3) = 9000e −0.3t
N ′(1) = 9000e −0.3(1) ≈ 6667
After 1 hour, the rate of change in the
number of informed people is increasing by
6667 per hour.
b. N ′(8) = 9000e −0.3(8) ≈ 816
After 8 hours, the rate of change in the
number of informed people is increasing by
816 per hour.
a.
Review Exercises for Chapter 4
38.
203
V (t ) = 50t 2 e −0.08t
V ′(t ) = 100te −0.08t + 50t 2 e −0.08t ( −0.08)
= 100te −0.08t − 4t 2 e −0.08t
We set V ′ (t ) = 0 to maximize V(t ) .
0 = 100te −0.08t − 4t 2 e −0.08t
0 = 4te −0.08t (25 − t )
A critical value is t = 25. Now we use the second derivative test.
V ′′(t ) = 100e −0.08t + 100te −0.08t ( −0.08) − 8te −0.08t − 4t 2 e −0.08t ( −0.08)
= 100e −0.08t − 8te −0.08t − 8te −0.08t + 0.32t 2 e −0.08t
= 100e −0.08t − 16te −0.08t + 0.32t 2 e −0.08t
V ′′(25) = 100e −2 − 400e −2 + 200e −2 = −100e −2 < 0
so V is maximized.
The present value is maximized in 25 years.
39.
p( x ) = 200e −0.25 x
a.
R ( x ) = xp( x ) = 200 xe −0.25 x
b.
R ′( x ) = 200e −0.25 x + 200 xe −0.25 x ( −0.25)
= 200e −0.25 x (1 − 0.25 x )
R ′( x ) = 0 when x = 1 = 4, which is the only critical value.
0.25
−0.25 x
′′
R ( x ) = 200e
( −0.25)(1 − 0.25 x ) + 200e −0.25 x ( −0.25)
= 200e −0.25 x ( −0.25)(1 − 0.25 x + 1)
= −50e −0.25 x (2 − 0.25 x )
R ′′(4) = −50e −1 < 0 ,
so R ( x ) is maximized at x = 4 .
Thus, quantity x = 4000 and price p(4) = 200e −1 ≈ $73.58 maximize revenue.
40.
a.
R ( x ) = p ⋅ x = (5 − ln x ) x = 5 x − x ln x
b.
R ′( x ) = 5 − ln x − x 1 = 4 − ln x
x
′
R ( x ) = 0 when x = e4 , which is the only critical value.
R ′′( x ) = − 1 < 0 when x = e4
x
so R is maximized.
()
p( e4 ) = 5 − ln( e4 ) = 1
Thus, the quantity x = e4 and the price $1 maximize revenue.
© 2010 Brooks/Cole, Cengage Learning.
204
41.
Chapter 4: Exponential and Logarithmic Functions
To find the maximum consumer expenditure, solve E′ ( p) = 0 , where E ( p ) = pD( p ) .
E ( p ) = p(25,000e −0.02 p )
E ′( p ) = 25,000e −0.02 p + 25,000e −0.02 p ( −0.02) p
= 25,000e −0.02 p − 500 pe −0.02 p
E′ ( p) = 0 when p = 50 , which is the only critical value.
We use the second derivative test to show that p = 50 is a maximum.
E ′′( p ) = 25,000e −0.02 p ( −0.02) − 500e −0.02 p − 500 pe −0.02 p ( −0.02)
= −500e −0.02 p − 500e −0.02 p + 10 pe −0.02 p
= −1000e −0.02 p + 10 pe −0.02 p
E ′′(50) = −1000e −0.02(50) + 10(50)e −0.02(50)
= −1000e −1 + 500e −1 < 0
so E is maximized.
The price $50 maximizes consumer expenditure.
42.
43.
on [–5, 15] by [–5, 15]
The function has a relative maximum at about
(4, 4.69) and a relative minimum at (0, 0).
There are inflection points at about (2, 2.17)
and (6, 3.21).
44.
on [–5, 5] by [–5, 5]
The function has a relative maximum at about
(–0.72, 0.12) and a relative minimum at about
(0.72, –0.12). There are inflection points at
about (–0.43, 0.07) and (0.43, –0.07).
To find the maximum consumer expenditure, solve E′ ( p) = 0, where E ( p ) = pD( p ) .
E ( p ) = p(200e −0.0013 p ) = 200 pe −0.0013 p
E ′( p ) = 200e −0.0013 p + 200 pe −0.0013 p ( −0.0013)
= 200e −0.0013 p − 0.26 pe −0.0013 p
E′ ( p) = 0 when the price is about 769, which is the only critical value.
E ′′( p ) = −0.26e −0.0013 p − 0.26e −0.0013 p − 0.26 pe −0.0013 p ( −0.0013)
= −0.52e −0.0013 p + 0.000338 pe −0.0013 p
E ′′(769) = −0.52e −0.0013(769) + 0.000338(769)e −0.0013(769)
= −0.26e −0.0013(769) < 0
so E is maximized.
The price $769 maximizes consumer expenditure.
45.
R ( x ) = xp( x ) = x (20e −0.0077 x ) = 20 xe −0.0077 x
R ′( x ) = 20e −0.0077 x + 20 xe −0.0077 x ( −0.0077)
= 20e −0.0077 x − 0.154 xe −0.0077 x
R′ (x) = 0 when the quantity is about 130.
R ′′( x ) = 20e −0.0077 x ( −0.0077) − 0.154e −0.0077 x + 0.0012 xe −0.0077 x
R ′′(130) = −0.154e −0.0077(130) − 0.154e −0.0077(130) + 0.0012(130)e −0.0077(130)
= −0.154e −0.0077(130) < 0
so R is maximized.
The quantity 130 maximizes revenue.
© 2010 Brooks/Cole, Cengage Learning.
Review Exercises for Chapter 4
46.
205
G (t ) = 5 + 2e0.01t
0.01t
(0.01) 0.02e0.01t
d ln G (t ) = 2e
=
dt
5 + 2e0.01t
5 + 2e0.01t
0.2
If t = 20, then the relative rate of change is 0.02e 0.2 ≈ 0.0033 = 0.33%
5 + 2e
47.
For t = 10,
48.
E ( p) =
G ′(t ) 0.02e0.01(10) 0.02e0.1
=
=
≈ 0.0031 = 0.31%
G (t ) 5 + 2e0.01(10) 5 + 2e0.1
− pD ′( p ) − p( −4 p )
4 p2
=
=
D( p )
63 − 2 p 2 63 − 2 p 2
E (3) = 4 ⋅ 9 = 36 = 4
63 − 18 45 5
Since E(p) < 1, demand is inelastic. Raising prices will increase revenue.
49.
−1/ 2 ⎤
⎡
( −1)
200 p
p
− pD ′( p ) − p ⎣ 200(600 − p )
⎦
E ( p) =
=
=
=
1/ 2
D( p )
400(600
p
)
2(600
−
− p)
400(600 − p )
350
= 350 = 0.7
2(600 − 350) 500
Since E( p) < 1 , demand is inelastic. Raising prices will increase revenue.
E (350) =
50.
E ( p) =
− pD ′( p ) − p( −0.528 p −1.44 ) 0.528
=
=
= 0.44
D( p )
1.2
1.2 p −0.44
Demand is inelastic.
51.
Relative rate of change of P ( x ) = d ln P ( x ) .
dx
d ln P ( x ) =
0.04 + 0.006 x 2
dx
3.25 + 0.04 x + 0.002 x 3
At x = 9,
52.
0.04 + 0.006(9)2
≈ 0.104 = 10.4%
3.25 + 0.04(9) + 0.002(9)3
a.
b.
Since demand is elastic, the dealer should lower the
price.
c.
E (8.7) ≈ 1 ; thus at $8700 elasticity equals 1.
on [0, 17] by [0, 15]
− pD ′( p )
E ( p) =
D( p )
− p( −20 + 2 p − 0.09 p 2 )
=
200 − 20 p + p 2 − 0.03 p 3
0.09 p 3 − 2 p 2 + 20 p
=
200 − 20 p + p 2 − 0.03 p 3
The demand is elastic at $10,000 because
E (10) ≈ 1.29 .
© 2010 Brooks/Cole, Cengage Learning.