Student Solutions Manual

```Chapter 4
Exponential and Logarithmic
Functions
4.1
Exponential Functions
(d)
7.
1. Use your HP48G.
For e2 , punch 2, then ex , to get
(23 − 32 )11/7 = (8 − 9)11/7 = −1
(33 )(3−2 ) = 33−2 = 3
9.
e2 = 7.389.
11.
For the TI-85, press 2nd ex then 2 ENTER.
Similarly e−2 = 0.135, e0.05 = 1.051,
√
e−0.05 = 0.951, e0 = 1, e = 2.718, e = 1.649,
1
√ = 0.607.
e
(32 )5/2
52
1
= 52−3 =
53
5
2 1/2 5
= [(3 ) ] = 35 = 243
13. If P dollars is invested at an annual interest
rate r and interest is compounded k times per
year, the balance after t years will be
B(t) = P
3.
1+
r kt
k
dollars and if interest is compounded
continuously, the balance will be
B(t) = P ert dollars.
(a) If P = 1, 000, r = 0.07, t = 10, and k = 1,
then
0.07 10
B(10) = 1, 000 1 +
1
= 1, 000(1.07)10 = \$1, 967.15.
y = 3x and y = 4x .
The x−axis is a horizontal asymptote.
5. (a)
(b)
(c)
272/3 = (271/3 )2 = 32 = 9
(−128)3/5 = −18.379
82/3 +163/4 = (81/3 )2 +(161/4 )3 = 4+8 = 12
115
(b) If P = 1, 000, r = 0.07, t = 10, and k = 4,
then
0.07 40
B(10) = 1, 000 1 +
4
= \$2, 001.60.
116
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
(c) If P = 1, 000, r = 0.07, t = 10, and k = 12,
then
0.07 120
B(10) = 1, 000 1 +
12
= \$2, 009.66.
B(30)
(d) If P = 1, 000, r = 0.07, t = 10, and
interest is compounded continuously, then
15.
= (P )(2)2 = 4P
When t = 0 P0 = 5, 000, and when t = 10
P = 8, 000. 8, 000 = 5, 000e10k , so e10k =
P =
5, 000
5, 000
=
= \$3, 534.12.
(1.0175)20
1.4178
17. (a)
B(5) = P
1+
0.07
4
20
29.
9, 000
9, 000
P =
=
= \$6, 361.42
20
(1.0175)
1.4148
B(5) = 9, 000 e−(0.07)(5) = \$6, 342.19
19. (a) The population in t years will be
P (t) = 50e0.02t
million. Thus P (0) = 50 million.
(b) The population 30 years from now will be
0.02(30)
P (30) = 50e
0.6
= 50e
= 91.11
million.
f (x) = e
f (1) = ek = 20
f (2) = e2k = (ek )2 = (20)2 = 400
f (x) = A2kx ,
f (0) = A = 20, f (2) = 20 (22k ) = 40, so
22k = 2.
f (8)
= 5, 000(e10k )3
3
8
= 5, 000 ×
= 20, 480 bacteria
5
P (t) = P0 e−kt
When t = 0, P0 = 25, 000 copies and when
t = 1 P = 10, 000 copies.
2
25, 000e−k = 10, 000, so e−k = and
5 2
2
P (2) = 25, 000e−2k = 25, 000
= 4, 000
5
copies.
31. The population density x miles from the center
of the city is D(x) = 12e−0.07x thousand people
per square mile.
(a) At the center of the city, the density is
D(0) = 12 thousand people per square
mile.
(b) Ten miles from the center, the density is
kx
23.
P (30)
8
and
5
= 5, 000.
Thus
21.
2
P (t) = P0 ekt
27.
Thus
(b)
r 120
= (P ) 1 +
4
r 60
= (P ) 1 +
4
The money has quadrupled, which is not
surprising since 30 = (2)(15).
0.7
B(10) = 1, 000e = \$2, 013.75.
0.07 20
= 5, 000.
B(5) = P 1 +
4
r 60
B(15) = 2P = (P ) 1 +
4
25.
= 20 (28k ) = 20(22k )4
= 20(24 ) = 320.
D(10) = 12e−0.07(10) = 12e−0.7 = 5.959
that is 5,959 people per square mile.
33. Let Q(t) denote the amount of the radioactive
substance present after t years.
Since the decay is exponential and 500 grams
were present initially,
Q(t) = 500e−kt
4.1. EXPONENTIAL FUNCTIONS
Moreover, since 400 grams are present 50 years
4
later, 400 = Q(50) = 500e−50k or e−50k = .
5
The amount present after 200 years will be
Q(200)
= 500e−200k = 500(e−50k )4
4
4
= 500
= 204.8 grams
5
35. f (t) = e−0.2t toasters will still be in use after t
years.
(a) After t = 3 years, f (3) = e−0.6 = 0.5488.
(b) The fraction of toasters still working after
2 years is
f (2) = e−0.4 = 0.6703, so the fraction of
toasters failing is 1.0 − 0.6703.
Similarly the fraction failing after 3 years
will be 1 − f (3) = 1.0 − 0.5488.
The fraction of toasters failing during the
third year is
1.0 − 0.5488 − (1.0 − 0.6703) = 0.1215.
(c) The fraction of toasters in use before the
first year is e−0.1 , so the fraction of
toasters failing during the first year is
1.0 − e−0.1 = 0.0952.
37. Let k be the number of compounding periods
per year.
(a) (P )(1 + i)k = (P )(1 + re ) from which
re = (1 + i)k − 1
(b) (P )er = (P )(1 + re ) from which
re = er − 1
39. Let the period of investment be 1 year and the
principal \$100. 0.082 4
= 108.46
B(t) = 100 1 +
4
B(t) = 100e0.081 = 108.44
Thus 8.2% is the winner.
41. (a)
N (t) = N0 e−0.217t
Let t = 0 be 200 B.C. In 2000 A.D. t = 2.2.
N (2200) = 500e−0.217(2.2) = 310.2
117
(b) In 950 t = 0, so t = 1 in 1950.
N (1) = 210e−0.217(1) = 169
This prediction is very close to the actual
count of 167.
(c) Writing Exercise —
Answers will vary.
Pi
43.
M=
1 − (1 + i)−n
0.09
P = 150, 000, i =
= 0.0075, n = 360.
12
M=
150, 000(0.0075)
= \$1, 206.93
1 − (1.0075)−360
45. (a) If the loan of \$5,000 is amortized over 3
years, the monthly payment would be
5000 (1 + 0.12)
−36
= 161.75
In this case you would receive
\$1, 000 + (161.75)(36) = \$6, 823
which is
6, 823−1, 000−(36)(160) = 6, 823−6, 760 = \$63
less than the buyer’s way.
(b) Writing Exercise —
Answers will vary
1 n
47.
A(n) = 1 +
.
n
A(−1, 000) = (1 − 0.001)−1,000 = 2.71964,
A(−2, 000) = (1 − 0.0005)−2,000 = 2.71896,
− 50,000
1
A(−50, 000) = 1 −
= 2.71831.
50, 000
The values approach e ≈ 2.71828.
5 n/3
49.
A(n) =
2−
2n
5 10/3
A(10) =
2−
= 6.4584
20
100/3
5
A(100) =
2−
= 7.12 × 109
200
which suggests lim (2 −
n→+∞
5
) = +∞
2n
118
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.2
Logarithmic Functions
7. Let’s call the given expression A.
A = e3 ln 2−2 ln 5
3
2
= eln 2 −ln 5
1. Use your calculator. For the HP48G, first enter
the number a, then press the ln key. For the
TI-85, press LN and the number. This gives
(since v ln u = ln uv ).
ln 1 = 0
ln 2 = 0.6931472
ln 5 = 1.6094379
1
ln
= ln 0.2 = −1.6094379
5
Now, for ln en , enter n, then press the ex key,
followed by the ln key (HP48G) or enter LN
followed by en (TI-85). Thus
ln e1 = 1 and ln e2 = 2.
A = eln 8−ln 25
= eln(8/25)
u
(since ln = ln u − ln v).
v
8
A=
because of the inverse relationship
25
between ex and ln x.
is equivalent to
These results should not be surprising since the
natural logarithm and the exponential functions
are inverses. ln en = n.
ln 2 = 0.06x,
from which x =
For ln 0 and ln(−2)
the calculator displays an error signal (such as
the letter E or a flashing light on the HP48G or
a complex number – a pair of numbers in
parentheses – on the TI-85.) This is due to the
domain of the logarithmic function which is
x > 0.
(Ask yourself “what exponent of e will give −2
?”
Answer: ex = −2 is impossible since ex > 0 for
all real x.
Remember that “log” means “exponent”.)
3.
ln e3 = 3 ln e = 3 × 1 = 3
since ln uv = v ln u and ln e = 1.
5. Let’s give the expression a name, say A, so that
we can handle it.
A = eln 5
which can be written as
2 = e0.06x
9.
11.
ln 2
= 11.55.
0.06
3 = 2 + 5e−4x
1
= e−4x
5
(using arithmetic),
−4x = ln
1
= − ln 5
5
u
= ln u − ln v and ln 1 = 0), from
v
ln 5
which x =
= 0.402.
4
t
− ln x =
+C
13.
50
or
t
ln x = − − C.
50
Thus
(since ln
x = e−C−t/50 = (e−C )(e−t/50 )
ln A = ln 5.
A solution (the only solution) is A=5. Thus
eln 5 = 5.
Actually this result is immediate from the
inverse relationship between ex and ln x.
because ar+s = ar as (which) certainly applies
when a = e.
4.2. LOGARITHMIC FUNCTIONS
15.
1
(ln 16 + 2 ln 2)
3
1
(ln 16 + ln 22 )
3
1
ln[(16)(22 )]
3
1
ln 26 = ln(26 )1/3
3
ln 22 .
ln x =
=
=
=
=
119
After a certain time the investment will have
grown to B(t) = 2P at the interest rate of 0.06.
Thus
2P = P e0.06t
2 = e0.06t
ln 2 = 0.06t
Thus ln x = ln 4 which is valid when x = 4.
3x = e2
17.
is valid if the logarithms of both members are
taken.
and t =
21.
B(t) = P ert ,
23.
25.
where P is the initial investment and r is the
interest rate compounded continuously.
Since money doubles in 13 years,
2P = B(13) = P e13r
2 = e13r
ln 2 = 13r
ax+1 = b
if ln ax+1 = ln b
(x + 1) ln a = ln b
ln b
x =
−1
ln a
log2 x
25
ln 25
or ln x
log5 (2x)
57
ln 57
ln x
ln √
1
ab3
= 5
= x
= 5 ln 2 = ln x
= 3.4657
= 7
= 2x
= ln(2x) = ln 2 + ln x
= 7 ln 5 − ln 2 = 10.5729
=
3 1/2
0 − ln(ab )
1
= − ln(ab3 )
2
1
= − (ln a + ln b3 )
2
1
= − (ln a + 3 ln b)
2
1
= − (2 + 9) = −5.5.
2
ln 2
= 11.55 years.
0.06
29. The balance after t years is
ln 3x = ln e2 = 2 ln e = 2
x ln 3 = 2, or
2
x =
= 1.82
ln 3
19.
B(t) = P ert .
27.
ln 2
= 0.0533. Thus the annual interest
13
rate is 5.33%.
or r =
31.
Q(t) = Q0 e−kt
Q0
= Q(1, 690) = Q0 e−1,690k
2
1
e−1,690k =
2
1
−1, 690k = ln
2
ln 2
k =
1, 690
Since Q0 = 50, 5 = 50e−kt ,
1
1
e−kt =
, −kt = ln
= − ln 10,
10
10
ln 10
1, 690 ln 10
t=
=
≈ 5, 614 years
k
ln 2
33. The number of bacteria is
Q(t) = Q0 ekt .
120
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Now R(t) = R0 e−kt = 0.997R0
Since 6,000 bacteria were present initially,
Q0 = 6, 000 so that
ln 0.997 = −
kt
Q(t) = 6, 000e
Q(20) = 9, 000 = 6, 000e20k
3
20k = ln
2
The forged Rembrandt painting is only
approximately 25 years old.
In the year 2000 t = 2, 000 − 1, 640 = 360.
Let p be the percentage of 14 C left.
R(t) = pR0
360 ln 2
or p = 0.9574
ln p = −
5, 730
k ≈ 0.0203 and Q(t) = 6, 000e0.0203t .
Q(t) = 500 − Ae−kt
35.
ln 2
t or t = 24.8372 years.
5, 730
and
The original Rembrandt painting will contain
approximately 95.74 % of 14 C.
Q(0) = 300 = 500 − A.
Thus A = 200 and
Q(t) = 500 − 200e−kt ,
or 45% of the original 133 I should be
detected.
e−(ln 2/20.9)(25) = 0.4364
(b)
Q(6) = 410 = 500 − 200e−6k .
e−6k =
9
or k =≈ 0.1331
20
or 43.64% of 133 I remains in the thyroid.
43.64 − 41.3 = 2.34% of 133 I still remains
in the rest of the patient’s body.
It follows that
Q(t) = 500 − 200e−0.1331t .
37. ¿From the half-life of
1
R0
2
1
ln
2
14
T = Ta + (Td − Ta )(0.97)t
so 40 = 10 + (98.6 − 10)(0.97)t ,
30 = 88.6(0.97)t ,
30
t ln 0.97 = ln
or t = 35.55
88.6
= R0 e−5,730k
ln 2
5, 730
= R0 e−kt = 0.28R0
ln 2
ln 0.28 = −
t
5, 730
5, 730 ln 0.28
t = −
= 10, 523years.
ln 2
Now R(t)
The artifacts at the Debert site in Nova Scotia
are about 10,500 years old.
39. ¿From the half-life of
1
R0
2
1
ln
2
43.
C
= −5, 730k or k =
14
C
= R0 e−5,730k
= −5, 730k or k =
ln 2
5, 730
e−(ln 2/20.9)(24) = 0.45
41. (a)
The murder occurred around 1:30 a.m. on
Wednesday morning. Blohardt was in the
slammer, so Scélerat must have done it.
45. (a)
R
ln I
Ia
(b)
ln Ib
Ib
Ia
Ib
ln I
= 8.3
ln 10
= 8.3 ln 10 = 19.1115
= e19.1115 = 1.995 × 108
=
= 7.1 ln 10 = 16.3484
= e16.3484 = 1.2589 × 107
= 15.8472
times more intense.
4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
47. Let t denote the number of years after 1960. If
the population P (measured in billions) is
growing exponentially and was 3 billion in 1960
(when t = 0), then
P (t) = 3ekt .
57.
e−3.5x
1 + 257e−1.1x
Using a graphing utility and tracing, we find
x = −5.06.
3, 500e0.31x =
59.
ln(x + 3) − ln x = 5 ln(x2 − 4)
x+3
ln
= 0
x(x2 − 4)5
Since the population in 1975 (when t = 15) was
4 billion,
4
= P (15) = 3e15k or k =
For P (t)
=
ekt
=
x+3
= 1. Using a graphing utility we
x(x2 − 4)5
find x = 2.28.
1
4
ln
15 3
40 so 3ekt = 40
40
or t ≈ 135
3
or
61. k =
ln 2
= 0.015. After 24 hours
46.5
100e−0.015(24) = 69.77 mg
The population will reach 40 billion in the year
1960 + 135 = 2095.
will be left. The time required for the isotope
to decline to 25 mg is
ln 0.25
t=−
= 92.4 hours.
0.015
49. (a) 0.05 = e−3k , ln(0.05) = −3k or k = 0.999
0.01 = e−0.999x
ln(0.01)
= 4.61 m.
0.999
(b) Writing exercise —
Answers will vary.
or x = −
51. (a) λ =
ln 2
. Therefore
k
Q(t) = Q0 e−(ln 2/λ)t
(b)
Q0 e−(ln 2/λ)t = Q0 (0.5)kt
− ln 2
− ln 2
1
=
=
(ln 0.5)λ
= ln 2λ
λ
logb b
1
loga b =
=
logb a
logb a
k=
53.
4.3
Therefore loga b logb a = 1.
55. 10x and log10 x are reflections about y = x.
121
1.
Differentiation of
Logarithmic and
Exponential Functions
f (x)
= e5x
f 0 (x) = e5x
3.
2
f (x) = ex
d
(5x) = 5e5x
dx
+2x−1
d
(x2 + 2x − 1)
dx
2
= (2x + 2)ex +2x−1 .
2
f 0 (x) = ex
+2x−1
122
5.
7.
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
30 + 10e−0.05x
d
f 0 (x) = 0 + 10e−0.05x (−0.05x)
dx
= −0.5e−0.05x .
f (x)
=
f (x)
=
2
21.
f 0 (0) = 1 and f (0) = 0. An equation of the
tangent line is
6x
(x + 3x + 5)e
d
f 0 (x) = (x2 + 3x + 5) e6x
dx
d
+e6x (x2 + 3x + 5)
dx
= (6x2 + 20x + 33)e6x .
f (x) = xe−x
f 0 (x) = −xe−x + e−x
= e−x (1 − x)
y − 0 = (1)(x − 0) or y = x
23.
f 0 (x)
9.
=
x 2
(1 − 3e )
f (x)
=
f 0 (x)
d
(1 − 3ex )
dx
= −6ex (1 − 3ex ).
f 0 (1) = 0 and f (1) = e2 . An equation of the
tangent line is
y = e2
2(1 − 3ex )
=
√
e2x
x2
x2 (2e2x ) − e2x (2x)
2(x − 1)e2x
=
x4
x3
f (x) =
1/2
√
x2 ln x
25.
= e(3x)
f (x) = x2 ln x =
2
√
d
x
+
2x
ln
x
f 0 (x) = e 3x (31/2 x1/2 )
f 0 (x) =
dx
2
√
√
1
3
= e 3x 31/2
x−1/2 = √ e 3x .
1
2
2 3x
f 0 (1) = and f (1) = 0. An equation of the
2
tangent line is
13.
f (x) = ln x3 = 3 ln x
1 d
3
y − 0 = ( 12 (x − 1) or x − 2y − 1 = 0
f 0 (x) = 3
x= .
x dx
x
11. f (x)
15.
17.
19.
= e
3x
= x2 ln x
d
d
f 0 (x) = x2 (ln x) + ln x (x2 )
dx
dx
= x(1 + 2 ln x).
f (x)
f (x)
=
f 0 (x)
=
f (x)
=
f 0 (x)
=
=
=
√
3
e2x
27.
ln f (x)
f 0 (x)
f (x)
2x/3
=e
2
2√
3
e2x/3
=
e2x
3
3
x+1
ln
x−1
x−1 d
x+1
x + 1 dx x − 1
x − 1 −2
x + 1 (x − 1)2
−2
.
x2 − 1
f (x)
=
1
ln(3x − 5).
6
5
3
−
,
x + 2 6(3x − 5)
5
(x + 2)5
1
√
−
6
3x − 5 x + 2 2(3x − 5)
= 5 ln(x + 2) −
=
f 0 (x) =
29.
(x + 2)5
√
.
6
3x − 5
f (x) = (x + 1)3 (6 − x)2 (2x + 1)1/3 .
ln f (x) = ln[(x + 1)3 (6 − x)2 (2x + 1)1/3 ]
= ln(x + 1)3 + ln(6 − x)2
+ ln(2x + 1)1/3
= 3 ln(x + 1) + 2 ln(6 − x)
1
+ ln(2x + 1).
3
4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
Differentiating leads to
f 0 (x)
f (x)
f 0 (x)
(c)
3
x+1
2
2(−1)
+
,
+
6−x
3(2x + 1)
√
= (x + 1)3 (6 − x)2 3 2x + 1
3
2
2
−
+
.
x + 1 6 − x 3(2x + 1)
=
2
31.
f (x) = 2x
ln f (x) = x2 ln 2
f 0 (x)
= 2x ln 2
f (x)
f 0 (x)
2
2x
=
+1
x ln 2
C(x) = e0.2x
33.
C 0 (x) = 0.2e0.2x
(a)
(d) R0 (x) = C 0 (x) when
x + 3 ln(x + 3)
= 2x
(x + 3)2
or x = 0.1805 (obtained with a graphing
utility.)
2
(e)
2x = x +
x
√
or x = 2.
37. (a) The population t years from now will be
(c)
e
=
x
0.2x
A0 (x)
e
=
(0.2x − 1)
x2
= xe−3x
= e−3x (1 − 3x)
R(x)
R0 (x)
(d) R0 (x) = C 0 (x) when
or x = 0.2049
(e)
0.2e0.2x =
e0.2x
x
or x = 5.
C(x) = x2 + 2
C 0 (x) = 2x
A(x)
A0 (x)
million. Hence the rate of change of the
population t years from now will be
P 0 (t) = 50e0.02t (0.02) = e0.02t
and the rate of change 10 years from now
will be
P 0 (t) = e0.2 = 1.22
million per year.
e−3x (1 − 3x) = 0.2e0.2x
(b)
=
0.2x
A(x)
(a)
x ln(x + 3)
x+3
1
x
0
[(x + 3)
+ ln(x + 3)
R (x) =
(x + 3)2
x+3
−x ln(x + 3)]
x + 3 ln(x + 3)
=
(x + 3)2
R(x)
P (t) = 50e0.02t
(b)
35.
123
= x + 2x−1
= 1 − 2x−2
(b) The percentage rate of change t years from
now will be
0 0.02t P (t)
e
100
= 100
P (t)
50e0.02t
100
=
=2
50
% per year, which is a constant,
independent of time.
39. (a) The value of the machine after t years is
Q(t) = 20, 000e−0.4t
dollars. Hence the rate of depreciation
after t years is
Q0 (t) = 20, 000e−0.4t (−0.4) = −8, 000e−0.4t
124
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
and the rate after 5 years is
(a)
Q0 (5) = −8, 000e−2 = −\$1, 082.68 per year.
(b) The percentage rate of change t years from
now will be
0 Q (t)
−8, 000e−0.4t
100
= 100
= −40
Q(t)
20, 000e−0.4t
% per year, which is a constant,
independent of time.
41. (a) The first year sales of the text will be
47.
f (x) = 20 − 15e−0.2x
thousand copies when x thousand
complementary copies are distributed.
If the number of complimentary copies
distributed is increased from 10,000, that
is when x = 10, by 1,000, that is , ∆x = 1,
the approximate change in sales is
49.
(b) The actual change in sales is
∆f
= f (11) − f (10)
= (20 − 15e−2.2 ) − (20 − 15e−2 )
= 0.368 or 368 copies.
Q(t) = Q0 e−0.0015t
43.
1
(a)
= e−0.0015t or t = 462
2
Half of the ozone will be depleted in 462
years.
(b)
0.2 = e−0.0015t , t = −
ln 0.22
= 1, 073
0.0015
80% of the ozone will be depleted in 1,073
years.
45.
C(t) = 0.2te−t/2
−t/2
at t = 2. C(2) = 0.147 or 14.7%.
(b) We want t such that
C(t) = 0.3(0.147) = 0.0441
Using a graphing utility, we find t = 6.9
hours.
2x
f (x) =
x
x
x(2
) ln 2 − 2x
2x (ln 2x − 1)
=
f 0 (x) =
x2
x2
x ln x
f (x) = x log10 x =
ln 10
1
0
f (x) =
(1 + ln x)
ln 10
51. By definition, the percentage rate of change of
f with respect to x is
∆f = f 0 (10)∆x.
Since f 0 (x) = 3e−0.2x and ∆x = 1,
∆f = f 0 (10) = 3e−2 = 0.406 thousand or
406 copies.
1
C (t) = 0.2e
+ 0.2t − e−t/2
2
t
−t/2
= 0.2e
1−
=0
2
0
100
f 0 (x)
f (x)
d
f 0 (x)
ln f (x) =
, it follows that the
dx
f (x)
percentage rate of change can be written as
d
100 ln f (x).
dx
53. The population of the town x years from now
will be
p
P (x) = 5, 000 x2 + 4x + 19
Since
¿From problem 51, the percentage rate of
change x years from now will be
p
d
100 ln 5, 000 x2 + 4x + 19
dx
d
1
= 100 [ln 5, 000 + ln(x2 + 4x + 19)]
dx
2
100(x + 2)
=
x2 + 4x + 19
Hence the percentage rate of change 3 years
from now will be
100(3 + 2)
= 12.5 % per year
32 + 12 + 19
4.4. ADDITIONAL EXPONENTIAL MODELS
125
f (x) = (3.7x2 − 2x + 1)e−3x+2
55.
f 0 (−2.17) = −428, 640
y = g(t) = 5 − 3e−t .
7.
The line y = 5 is a horizontal asymptote.
4.4
Additional Exponential
Models
1.
y = f (t) = 2 + et
The line y = 2 is a horizontal asymptote.
9.
3.
2
.
1 + 3e−2t
The lines y = 0 and y = 2 are horizontal
asymptotes.
y = f (x) =
y = g(x) = 2 − 3ex .
The line y = 2 is a horizontal asymptote.
11.
f (x) = xex
f 0 (x) = ex (x + 1) = 0
when x = −1.
f 00 (x) = ex (1 + x + 1)
= ex (x + 2) = 0
5.
y = f (x) = 3 − 2(2−x ).
The line y = 3 is a horizontal asymptote.
when x = −2.
(−1, −0.37) is a minimum,
(−2, −0.27) is a point of inflection.
126
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Note that f (x) = xex < 0 if x < 0 and
decreasing (if x < −1).
The line y = 0 is an upper bound (the curve is
never above this line for x < 0.)
This strongly suggests that the x axis is a
horizontal asymptote.
x
−∞
−2
−1
∞
f (x)
0
−0.27 −0.37 ∞
f 0 (x)
−
−
0
+
f 00 (x)
−
0
+
+
f (x) = x2 e−x .
15.
f 0 (x) = e−x [−x2 + 2x] = xe−x (2 − x) = 0
when x = 0, 2.
f 00 (x)
13.
f (x)
f 0 (x)
= e−x [−2x + 2 − (−x2 + 2x)]
= e−x (x2 − 4x + 2) = 0
√
4 ± 16 − 8
,
when x =
2
so x = 0.59 or x = 3.41.
(0, 0) is a minimum,
(2, 0.54) is a maximum,
(0.59, 0.19) and (3.41, 0.38) are points of
inflection.
Note that f (x) = x2 e−x > 0 if x > 0 and
decreasing.
The line y = 0 is a lower bound (the curve is
never below this line.)
This strongly suggests that the x axis is a
horizontal asymptote.
= xe2−x
= e2−x (1 − x) = 0
when x = 1.
f 00 (x)
= e2−x [−1 + (1 − x)(−1)]
= e2−x (x − 2) = 0
when x = 2.
(1, 2.7) is a maximum,
(2, 2) is a point of inflection.
Note that f (x) = xe2−x > 0 if x > 0 and
decreasing for x > 1.
The line y = 0 is a lower bound (the curve is
never below this line if x > 0.)
This strongly suggests that the x axis is a
horizontal asymptote.
x
−∞
1
2 ∞
f (x) −∞ 2.7 2
0
0
f (x)
+
0
− −
f 00 (x)
−
−
0 +
x −∞ · · · 0 · · · 0.59
2 3.41 ∞
f (x)
∞ · · · 0 · · · 0.19 0.5 0.38 0
f 0 (x)
− − 0 +
+
0
− −
f 00 (x)
+ + + +
0 −
0 +
17.
4.4. ADDITIONAL EXPONENTIAL MODELS
f (x)
=
f 0 (x)
=
=
f 00 (x)
=
=
6
= 6(1 + e−x )−1 .
1 + e−x
−6(1 + e−x )−2 (−e−x )
6
> 0.
ex (1 + e−x )2
6
2x
e (1 + e−x )4
[0 − (ex )(2)(1 + e−x )(−e−x )
−(1 + e−x )2 ex ]
6(1 − e−x )
− x
=0
e (1 + e−x )3
when e−x = 1 or x = 0.
(0, 3) is a point of inflection.
Note that if x → −∞ 1 + e−x → ∞ and y → 0,
so the x axis is a horizontal asymptote.
Similarly if x → +∞ 1 + e−x → 1 and y → 6,
so the line y = 6 is a horizontal asymptote also.
When x < 0, f (x) is concave up, and
when x > 0, f (x) is concave down.
19.
f (x)
f 0 (x)
= (ln x)2
ln x
= 2
=0
x
when x = 1.
f 00 (x) =
2
(1 − ln x) = 0
x2
when x = e.
(1, 0) is a minimum,
(e, 1) is a point of inflection.
Note that the curve seems to level off to the
right, but this is an optical illusion.
f (x) will keep increasing beyond all bounds.
For example f (1, 000) = 47.7 and
f (1098 ) = 51, 964.
127
The y–axis is a vertical asymptote.
When x < 1, f (x) is decreasing.
When x > 1, f (x) is increasing.
When x < e, f (x) is concave up.
When x > e, f (x) is concave down.
21. (a) The reliability function is
f (t) = 1 − e−0.03t .
As t increases without bound, e−0.03t
approaches 0 and so f (t) approaches 1.
Furthermore, f (0) = 0.
The graph is like that of a learning curve.
(b) The fraction of tankers that sink in fewer
than 10 days is f (10) = 1 − e−0.3 .
The fraction of tankers that remain afloat
for at least 10 days is therefore
1 − f (10) = e−0.3 = 0.7408.
(c) The fraction of tankers that can be
expected to sink between the 15th and
20th days is f (20) − f (15)
= (1 − e−0.6 ) − (1 − e−0.45 )
= −e−0.6 + e−0.45
= −0.5488 + 0.6373 = 0.0888.
23. The temperature of the drink t minutes after
leaving the refrigerator is
f (t) = 30 − Ae−kt .
128
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Since the temperature of the drink when it left
the refrigerator was 10 degrees Celsius,
f (x) = 15 − 20e−0.3x
29.
x = 9 thousand.
f (9) = 15 − 20e−2.7 = 13.656
10 = f (0) = 30 − A or A = 20.
Thus
−kt
f (t) = 30 − 20e
.
(a)
f 0 (9) = 6e−2.7 = 0.403 thousand
(b)
f (10) = 15 − 20e−3 = 14.004
Since the temperature of the drink was 15
degrees Celsius 20 minutes later,
15 = f (20) = 30 − 20e−20k or e−20k =
3
.
4
The temperature of the drink after 40 minutes
is therefore
f (40)
25. (a)
= 30 − 20e−40k = 30 − 20(e−20k )2
2
3
= 30 − 20
= 18.75 degrees.
4
f (t) =
The actual increase was
14.004 − 13.656 = 0.348 or 348 books. The
estimate of 403 books was not too bad an
approximation.
Q(t) = 80(4 + 76e−1.2t )−1
31.
Q0 (t)
= 80(−1)(4 + 76e−1.2t )−2 (76)e−1.2t (−1.2)
e−1.2t
= 80 × 76 × 1.2
(4 + 76e−1.2t )2
After 2 weeks (at the end of the second week)
2
.
1 + 3e−0.8t
Q(2)
=
Q00 (t)
(b) f (0) = 0.5 thousand people (500 people).
2
(c) f (3) =
= 1.572, so 1,572
1 + 3 × 0.0907
people have caught the disease.
(d) The highest number of people who can
2
contract the disease is
= 2 or 2,000
1+0
people.
Note that the graph is shown only for
t ≥ 0.
27. Q(t) = 40 − Ae−kt
Q(0) = 20, 20 = 40 − A so that A = 20.
1
Q(1) = 30, 30 = 40 − 20e−k and e−k = .
2
Q(3) = 40 − 20e−3k = 37.5 units per day.
=
e−2.4
(4 + 76e−2.4 )2
5.576 or 5,576 people
= 80 × 76 × 1.2
7296(−1.2)(4 + 76e−1.2t − 152e−1.2t )
e1.2t (4 + 76e−1.2t )3
which equals 0 if 76e−1.2t = 4 or t = 2.45.
The disease is spreading most rapidly
approximately 2 12 weeks after the outbreak.
Cekt
33.
P (t) =
1 + Cekt
(a)
P (0) = P0 =
or C
(b)
=
C
1+C
P0
1 − P0
Cekt
t→∞ 1 + Cekt
C
= lim −kt
= 1 = 100%
t→∞ e
+C
lim
35. (a) The profit per VCR is \$x − 125. The
number of units sold is 1, 000e−0.02x . The
weekly profit is
P (x) = 1, 000(x − 125)e−0.02x
4.4. ADDITIONAL EXPONENTIAL MODELS
129
200
200(6 + t)
1
=
6+t
which will be equal to the prevailing interest
1
= 0.08 or
rate of 8% when
6+t
1
t=
− 6 = 6.5.
0.08
1
Moreover,
> 0.08 when 0 < t < 6.5
6+t
1
< 0.08 when 6.5 < t.
and
6+t
Hence the percentage rate of growth of the
value of the collection is greater than the
prevailing interest rate when 0 < t < 6.5
and less than the prevailing interest rate when
t > 6.5. Thus the collection should be sold in
6.5 years.
c
y=
(e−at − e−bt )
41.
b−a
c
(a)
y0 =
(−ae−at + be−bt ) = 0
b−a
a
when ae−at = be−bt or = e(a−b)t
b
a
1
Thus t =
ln
.
b
a−b
In the long run both exponential terms
approach zero, so y → 0.
=
(b)
P 0 (x)
=
1, 000[−0.02(x − 125)e−0.02x
+e−0.02x ] = 0
when 0.02(x − 125) = 1 or x = 175.
37. The percentage rate of change of the market
price
√
V (t) = 8, 000e t
of the land (expressed in decimal form) is
V 0 (t)
V (t)
√
=
8, 000e t 1
1
√
√ = √
t
2 t
2 t
8, 000e
which will be equal to the prevailing interest
rate of 6 % when
2
1
1
√ = 0.06 or t =
= 69.44.
0.12
2 t
1
Moreover, √ > 0.06 when 0 < t < 69.44 and
2 t
1
√ < 0.06 when 69.44 < t.
2 t
Hence the percentage rate of growth of the
value of the land is greater than the prevailing
interest rate when 0 < t < 69.44 and less than
the prevailing interest rate when 69.44 < t.
Thus the land should be sold in 69.44 years.
39. Since the stamp collection is currently worth
\$1,200 and its value is increases linearly at the
rate of \$200 per year, its value t years from now
is
V (t) = 1, 200 + 200t.
The percentage rate of change of the value
(expressed in decimal form) is
V 0 (t)
V (t)
=
200
1, 200 + 200t
(b)
N (t) =
43. (a)
500(0.03)(0.4)
t
0
N (0)
= 500(0.03)(0.4) = 15
N (5)
= 500(0.03)(0.4) ≈ 482 employees.
5
t
500(0.03)(0.4)
ln 0.6
if (0.4)t =
= 0.145677
ln 0.03
ln 0.145677
t =
= 2.10 years.
ln 0.4
0
lim N (t) = 500(0.03) = 500 employees.
300
t→∞
=
130
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
(b)
F (t)
=
500(0.03)−(0.4)
−t
f (x) = x(e−x + e−2x )
49.
N (t) is bounded between 0 and 500 while 500 is
the lower bound for F (t) and there is no upper
bound.
lim f (x) = 0
x→∞
Note: The portion of the graph in the third
quadrant is likely to be hidden from view on
your graphing utility unless you specificaly
request a negative domain.
The high point occurs at (0.76, 0.52).
45. Let’s assume continuous growth, so
Q(t) = Q0 e0.06t
Let t = 0 be 1947. Q0 = 1, 139.
(a)
0.06(7)
Q(7) = 1, 139e
Q(53) = 27, 389.3
√
= 1, 733.5
(b) We want 2, 000 = 1, 139e0.06t or
1.756 = e0.06t .
This leads to t = 9.38.
2, 278 = 1, 139e0.06t or 2 = e0.06t .
This leads to t = 11.55.
(c) Writing Exercise —
Answers will vary.
P (x) = λ2 xe−λx , 0 < λ < e
47.
(a)
(b)
P 0 (x)
= λ2 e−λx + λ2 x(−λe−λx )
= λ2 e−λx (1 − xλ) = 0
1
1
λ
at x = . P
= λe−1 =
λ
λ
e
P (t) = 20, 000te
51.
0.4t−0.07t
A graphing utility indicates a maximum present
value of \$150, 543 at t = 44.38 years.
4.5
Review Problems
Review Problems
1. (a)
If f (x) = 5e−x ,
then f (x) approaches 0 as x increases
without bound, and f (x) increases without
bound as x decreases without bound. The
y–intercept is f (0) = 5.
4.5. REVIEW PROBLEMS
131
If f (x) = 5 − 2e−x ,
(b)
As x decreases without bound, e2x
approaches 0 and thus f (x) approaches 2.
5
3+2
The y−intercept is f (0) =
= .
1+1
2
Applying the quotient rule to determine
f 0 (x) yields
then f (x) approaches 5 as x increases
without bound, and f (x) decreases
without bound as x decreases without
bound. The y–intercept is f (0) = 3.
f 0 (x) =
(c)
If f (x) = 1 −
6
,
2 + e−3x
the y–intercept is
6
6
f (0) = 1 −
=1−
= −1.
0
2+e
2+1
As x increases without bound, e−3x
approaches 0, and so
lim f (x) = 1 −
x→∞
2e2x
.
(e2x + 1)2
Since 2e2x > 0, there are no critical points
and f (x) increases for all x.
To find f 00 (x), differentiate f 0 (x) using the
quotient rule obtaining
6
= −2.
2+0
f 00 (x) =
4e2x (1 − e2x )
(e2x + 1)3
There is one inflection point
5
(0, f (0)) = (0, ) since f 00 (0) = 0. The
2
function f (x) is concave down for x > 0
and concave up for x < 0.
As x decreases without bound, e−3x
increases without bound, and so
lim f (x) = 1 − 0 = 1.
x→−∞
2. (a)
If f (x) = Ae−kx
and f (0) = 10, then 10 = Ae0 .
Hence f (x) = 10e−kx .
Since f (1) = 25, 25 = 10e−k or e−k =
(d)
If f (x) =
3 + 2e−2x
,
1 + e−2x
then f (x) approaches 3 as x increases
without bound, since e−2x approaches 0.
Rewriting the function, by multiplying
numerator and denominator by e2x yields
f (x) =
3e2x + 2
.
e2x + 1
5
.
2
Then f (4) = 10e−4k = 10(e−k )4
4
5
= 10
= 390.625.
2
(b) If f (x) = Aekx and f (1) = 3 as well as
f (2) = 10,
then 3 = Aek and 10 = Ae2k , two
equations in two unknowns.
3
Aek
Division eliminates A, so
=
,
10
Ae2k
k
3
e
10
= 2k , or ek =
.
10
e
3
132
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
10
10
Since e =
,3=A
3
3
9
and so A =
. Thus
10
9 k 3
100
9 3k
f (3) =
e =
(e ) =
10
10
3
k
(c) If
and f (0) = 50, then 50 = 30 + Ae0 or
A = 20.
Hence, f (x) = 30 + 20e−kx .
Since f (3) = 40, 40 = 30 + 20e−3k ,
1
10 = 20e−3k , or e−3k = .
2
(d)
V (10) = 50, 000e2 ln(2/5)
= 50, 000eln(4/25)
4
= \$8, 000.
= 50, 000
25
4. The sales function is
f (x) = 30 + Ae−kx
Thus f (9)
Hence V (t) = 50, 000e[1/5 ln(2/5)]t and so
= 30 + 20e−9k
= 30 + 20(e−3k )3
3
1
= 30 + 20
= 32.5.
2
6
1 + Ae−kx
6
and f (0) = 3 then 3 =
, or A = 1.
1 + Ae0
6
Hence, f (x) =
.
1 + e−kx
6
Since f (5) = 2, 2 =
,
1 + e−5k
−5k
−5k
2 + 2e
= 6, or e
= 2. Then,
Q(x) = 50 − 40e−0.1x
units, where x is the amount (in thousands)
spent on advertising.
(a) As x increases without bound, Q(x)
approaches 50. The vertical axis intercept
is Q(0) = 10. The graph is like that of a
learning curve.
If f (x) =
f (10)
=
=
6
6
=
−10k
1+e
1 + (e−5k )2
6
6
= .
2
1 + (2)
5
3. Let V (t) denote the value of the machine after t
years.
Since the value decreases exponentially and was
originally \$50,000, it follows that
V (t) = 50, 000e−kt .
Since the value after 5 years is \$20,000,
20, 000
−5k
e
or k
−5k
= V (5) = 50, 000e
2
=
,
5
1 2
= − ln .
5 5
(b) If no money is spent on advertising, sales
will be Q(0) = 10 thousand units.
(c) If \$8,000 is spent on advertising, sales will
be Q(8) = 50 − 40e−0.8 = 32.027 thousand
or 32,027 units.
(d) Sales will be 35 thousand if Q(x) = 35,
that is, if
50 − 40e−0.1x
= 35,
3
e−0.1x =
,
8
ln(3/8)
x = −
= 9.808
0.1
thousand or \$9,808.
,
(e) Since Q(x) approaches 50 as x increases
without bound, the most optimistic sales
projection is 50,000 units.
4.5. REVIEW PROBLEMS
133
5. The output function is
Q(t) = 120 − Ae−kt .
Since Q(0) = 30, 30 = 120 − A or A = 90.
Since Q(8) = 80, 80 = 120 − 90e−8k ,
4
−40 = −90e−8k or e−8k = . Hence,
9
Q(4)
= 120 − 90e−4k = 120 − 90(e−8k )1/2
1/2
4
= 120 − 90
= 60 units.
9
6. The population t years from now will be
P (t) =
30
.
1 + 2e−0.05t
(a) The vertical axis intercept is
30
P (0) =
= 10 million.
1+2
As t increases without bound, e−0.05t
approaches 0. Hence,
lim P (t)
t→∞
=
lim
t→∞
30
= 30
1 + 2e−0.05t
As t decreases without bound, e−0.05t
increases without bound.
Hence, the denominator 1 + 2e−0.05t
increases without bound and
30
lim P (t) = lim
= 30.
t→∞
t→∞ 1 + 2e−0.05t
(b) The current population is P (0) = 10
million.
(c) The population in 20 years will be
P (20)
=
=
30
1 + 2e−0.05(20)
30
= 17.2835
1 + 2e−1
million or 17,283,500 people.
(d) In the long run (as t increases without
bound), e−0.05t approaches 0 and so the
population P (t) approaches 30 million.
7. (a)
ln e5 = 5 since n = ln en .
(b)
eln 2 = 2 since n = eln n .
(c)
(d)
8. (a)
e3 ln 4−ln 2
3
= eln 4 −ln 2
64
ln
= e 2 = eln 32 = 32.
ln(9e2 ) + ln(3e−2 ) = ln[(9e2 )(3e−2 )]
= ln 27 = 3 ln 3.
8 = 2e0.04x ,
e0.04x = 4,
0.04x = ln 4
x = 34.657.
(b)
= 1 + 4e−6x ,
4e−6x = 4
−6x = ln 1 = 0, or x = 0.
(c)
4 ln x = 8, ln x = 2,
or x = e2 = 7.389.
(d)
5x = e3 ,
ln 5x = ln e3 ,
x ln 5 = 3, or
3
x =
= 1.864.
ln 5
5
9. Let Q(t) denote the number of bacteria after t
minutes. Since Q(t) grows exponentially and
5,000 bacteria were present initially,
Q(t) = 5, 000ekt .
Since 8,000 bacteria were present after 10
minutes,
8
8, 000 = Q(10) = 5, 000e10k , e10k = , or
5
1
8
k=
ln .
10 5
134
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
The bacteria will double when
= t log3 t2 ,
2
t ln t
=
ln 3
2
(1 + ln t).
g 0 (t) =
ln 3
(f )
g(t)
Q(t) = 10, 000,
that is, when
5, 000ekt
kt
or t
10. (a)
(b)
=
=
10, 000
11. (a) Using the formula
ln 2
ln 2
10 ln 2
r kt
=
=
1
+
B(t)
=
P
k
ln(8/5)
k
= 14.75 or 14. min. and 45 seconds.
with P = 2, 000, B = 5, 000, r = 0.08, and
k = 4,
f (x) = 2e3x+5 ,
0.08 4t
5,
000
=
2,
000
1
+
0
3x+5 d
f (x) = 2e
(3x + 5)
4
dx
3x+5
5
= 6e
.
(1.02)4t =
2
= x2 e−x ,
d
d
f 0 (x) = e−x (x2 ) + x2 e−x
dx
dx
= x(2 − x)e−x .
f (x)
B(t) = P ert
(c)
(d)
(e)
1 ln(5/2)
= 11.57 years.
4 ln 1.02
(b) Using the formula
or t =
p
g(x) = ln x2 + 4x + 1
1
ln(x2 + 4x + 1),
=
2
1
1
g 0 (x) =
2 x2 + 4x + 1
d
(x2 + 4x + 1)
dx
x+2
=
.
2
x + 4x + 1
= x ln x2 = 2x ln x,
d
d
0
h (x) = 2 x ln x + ln x x
dx
dx
= 2(1 + ln x).
h(x)
f (t)
=
t
,
ln 2t
(ln 2t)(1) − t
0
f (t)
=
=
with P = 2, 000, B = 5, 000, and r = 0.08,
5, 000 = 2, 000e0.08t ,
e0.08t =
or t =
5
,
2
ln(5/2)
= 11.45.
0.08
12. Compare the effective interest rates.
The effective interest rate for 8.25 %
compounded quarterly is
rk
0.0825 4
1+
−1 =
1+
−1
k
4
= 0.0851 or 8.51 %.
The effective interest rate for 8.20 compounded
continuously is
(ln 2t)2
ln 2t − 1
.
(ln 2t)2
1
2t
(2)
er − 1 = e0.082 − 1 = 0.0855 or 8.55 %.
13. (a) Using the present value formula
r −kt
P =B 1+
k
4.5. REVIEW PROBLEMS
135
with B = 2, 000, t = 10, r = 0.0625, and
k = 12,
0.0625 -120
P = 2, 000 1 +
12
2, 000
=
= \$1, 072.26.
1.8652182
(b) Using the present value formula
P = Be−rt
(a) The rate of change of the carbon monoxide
level t years from now is Q0 (t) = 0.12e0.03t ,
and the rate two years from now is
Q0 (2) = 0.12e0.06 = 0.13 parts per million
per year.
(b) The percentage rate of change of the
carbon monoxide level t years from now is
0 Q (t)
.12e.03t
100
= 100
= 3% per year
Q(t)
4e.03t
which is a constant, independent of time.
with B = 2, 000, t = 10, and r = 0.0625,
P = 2, 000e−0.0625(10) = \$1, 070.52.
14. (a)
1+
0.0625
2
2 × 10
8, 000
=
(P )
P
=
8, 000
= 4, 323.25
(1.03125)20
(b)
15.
18. Let F (p) denote the profit, where p is the price
per camera. Then
F (p) =
(number of cameras sold)
(profit per camera)
= 800(p − 40)e−0.01p .
0
F (p) = 800[e−0.01p (1)
+(p − 40)e−0.01p (−0.01)]
= 8e−0.01p (140 − p) = 0
8, 000 = P e0.0625×10
P = 8, 000e−0.625 = 4, 282.09
B(t)
=
r
=
2, 054.44 = 1, 000e12r
ln 2.05444
= 0.06, or r = 6%
12
16. At 6% compounded annually, the effective
interest rate is
rk
0.06 1
1+
−1 =
1+
−1
k
1
= 0.06.
At r % compounded continuously, the effective
interest rate is er − 1. Setting the two effective
rates equal to each other yields
er − 1 = 0.06,
when p = 140.
Since F 0 (p) > 0 (and F is increasing) for
0 < p < 140,
and F 0 (p) < 0 (and F is decreasing) for
p > 140,
it follows that F (p) has its absolute maximum
at p = 140.
Thus the cameras should be sold for \$140
apiece to maximize the profit.
19. (a)
f (x) = xe−2x ,
f 0 (x) = xe−2x (−2) + e−2x (1)
= e−2x (1 − 2x) = 0 whenx =
1
f
2
=
f 00 (x) =
1
.
2e
4e−2x (x − 1) = 0
er = 1.06, r = ln 1.06 = 0.0583 or 5.83 %.
when x = 1. f (1) =
17. The average level of carbon monoxide in the air
t years from now is Q(t) = 4e0.03t parts per
million.
1
.
e2
1
.
2
136
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4e−x
> 0.
(1 + e−x )2
1
f 00 (x) =
[(1 + e−x )2
(1 + e−x )4
(4e−x )(−1) − 4e−x (2)
(1 + e−x )(e−x )(−1)]
4e−x (e−x − 1)
=
= 0 when x = 0
(1 + e−x )3
1
1
∞
2
1
1
f (x) −∞
0
2e
e2
0
f (x)
∞ +
0 − − − 0
↑
↓
↓ ↓
f 00 (x)
− − −
0 +
↓
↓ ↓
↑
1 1
,
is the absolute maximum
2 2e
1
while 1, 2 is a point of inflection.
e
x −∞
=
−∞
0
1
x
f (x)
f 0 (x)
0
2
+
↑
0
+
+
↑
↑
f 00 (x)
+
−
up
↓
(0, 2) is a point of inflection.
lim
x→−∞
(b)
f (x)
f 0 (x)
f 00 (x)
x
∞
4
.0707
4
=0
1 + e−x
because the denominator increases beyond
all bounds.
−x
= e −e
= ex + e−x > 0
= ex − e−x = 0 when x = 0
lim
x→∞
4
=4
1 + e−x
because e−x → 0.
So y = 0 and y = 4 are horizontal
asymptotes.
There are no extrema.
x −∞
0
∞
f (x) −∞
0
∞
f 0 (x)
∞ + + +
↑ ↑ ↑
f 00 (x)
− 0 +
↓
↑
(0, 0) is a point of inflection.
(d)
f (x) =
f 0 (x) =
ln(x2 + 1)
2x
= 0 when x = 0
x2 + 1
f (0) = 0.
(c)
f (x)
=
f 0 (x)
=
4
= 4(1 + e−x )−1 ,
1 + e−x
d
4(−1)(1 + e−x )−2 (1 + e−x )
dx
f 00 (x) =
2(1 − x)(1 + x)
=0
(x2 + 1)2
4.5. REVIEW PROBLEMS
137
when x = ±1. f (±1) = ln 2.
x −∞
−1
0
1
f (x)
∞
ln 2
0
ln 2
f 0 (x)
0 −
− − 0 +
+
↓
↓ ↓
↑
↑
f 00 (x)
0 −
0 + 2 +
0
↓
↑
↑
(0, 0) is the absolute minimum
while (±1, ln 2) are points of inflection.
aR0 = R0 e−kt , a = e−15,000k .
∞
∞
+ 0
↑
− 0
↓
22.
1
Since the half-life is 5,730, = e−5,730k or
2
ln 2
k=
and
5, 730
a = e−15,000 ln 2/5,730 = e−1.8145 = 0.1629.
(a)
R(t) = R0 e−(ln 2)t/5,730
R(1, 960) = R0 e−(ln 2)(1,960)/5,730
= R0 e−0.2371 = 0.7889R0
Thus about 78.89% of
in the shroud.
(b) Since 92.3% of
√
t
√
=
2, 000e t
√
2, 000e t
1
√
2 t
1
= √
2 t
which will be equal to the prevailing interest
1
rate of 7 % when √ = 0.07 or
2
t
2
1
t=
= 51.02.
0.14
1
Moreover, √ > 0.07 when 0 < t < 51.02
2 t
1
and √ < 0.07 when 51.02 < t.
2 t
Hence the percentage rate of growth of the
collection is greater than the prevailing interest
rate when 0 < t < 51.02 and less than the
prevailing interest rate when 51.02 < t.
Thus the coin collection should be sold in 51.02
years.
21. Let a be the desired ratio. Then
R(t) = R0 e−kt ,
14 C
1988 − 662 = 1326 so Pierre d’Arcis’s
suspicions were well founded.
.
Hence, the percentage rate of change of the
value of the collection (expressed in decimal
form) is
V 0 (t)
V (t)
should be left
is left in the forgery,
(ln 2)t
0.923 = exp −
5, 730
(ln 0.923)(5, 730)
t = −
= 662
ln 2
20. The value of the coin collection in t years is
V (t) = 2, 000e
14 C
f (t) = 70 − Ae−kt
23.
f (0) = 70 − A = 212 or A = −142. Thus
f (t) = 70 + 142e−kt . Let T0 be the ideal
temperature. Then
T0 + 15 = 70 + 142e−k(5) and
T0 = 70 + 142e−k(7)
Solving (with the SOLVE utility of our
calculator) we get k = 0.091 and T0 = 1450 .
24. (a)
D(t) = (D0 − 0.00046)e−0.162t + 0.00046
With D0 = 0.008,
D(10)
= (0.008 − 0.00046)e−1.62 + 0.00046
= 0.00195
and D(25) = 0.00590.
lim D(t) = 0
(b)
t→∞
138
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
25. (a) To compute doubling time in terms of the
rate set 2P = P ert and solve for t.
2 = ert or t =
ln 2
r
P (5, 000) = e−(ln 2)5,000/5,730 = 54.62% ≈ 55%
In the chart below find a comparison of
the rules of 69, 70, 72 with the true
doubling time.
r
0.04 0.06 0.09 0.10
69 17.25 11.50 7.67 6.9
70 17.5 11.67 7.78 7.0
72
18
12
8 7.2
True 17.33 11.55 7.7 6.93
28. The Bronze age began about 5,000 years ago
(around 3,000 B.C.).
The maximum percentage is
29. (a)
0.12
5.75
5.83
6
5.78
69
The rule of 69 is closest because
is
100
closest to ln 2.
(b)
A(x)
=
A0 (x)
=
ln x − 2
110
, for 10 ≤ x.
x
3 − ln x
110
=0
x2
when x = e3 = 20.0855 years. A(10) = 3.3284
so a person’s aerobic capacity is maximized at
about age 20.
27.
30.
f 0 (x)
=
4e−(ln x)
√
for x > 0
πx
−(ln x)2
=
4 e
√
π x2
[−2 ln x − 1] = 0
1
when x = √ = 0.6065.
e
According to the graphing utility, the most
common age is at (0.6065, 2.8977).
= Cke−kt
100k
=
ekt − 1
T (t) = 35e−0.32t
27 = 35e−0.32t or t = 0.811 min.
Rescuers have about 49 seconds before the girl
looses consciousness.
dT
= −35(0.32)e−0.32t
dt
27
35
dT
27
= (−35)(0.32)
= −8.64
dt
35
At T = 27 or e−0.32t =
2
f (x)
df
df
100
f
(c) Writing Exercise —
Answers will vary.
(b) Writing Exercise —
Answers will vary.
26.
f (t) = C(1 − e−kt ) = 0.008C
0.992 = e−2k or k = .00402
Thus the girl’s temperature is dropping at 8.64
degrees per minute.
31. (a)
y
y0
= 0.125(e4x + e−4x )
= 0.5(e4x − e−4x ) = 0
when e8x = 1 at x = 0.
4.5. REVIEW PROBLEMS
139
f (t) = 30 − Ae−kt
34.
Solve for −Ae−kt = f (t) − 30.
f 0 (t) = −Ae−kt (−k)
= kAe−kt = k[30 − f (t)]
where k is a constant of proportionality and
f (t) is the temperature of the drink.
P H = − log10 [H3 O+ ]
35.
(b) Writing exercise —
Answers will vary.
32.
k1
k2
k1
k2
k1
ln
k2
33. (a)
For milk and lime, P Hm = 3P Hl .
For lime and orange, P Hl = 0.5P Ho .
E0
= A exp −
RT1
E0
= A exp −
RT2
E0
E0
= exp
−
RT2
RT1
E0 1
1
=
−
R T2
T1
2 t
V (t) = V0 1 −
L
3.2
= 1.6
2
[H3 O+ ]l = 10−1.6 = 0.0251
202.31
P (t) =
36.
1 + e3.938−0.314t
Hint: Use the equation writer on the HP48G to
enter the function.
For the TI-85, press 2nd CALC
EVALF(202.31/(1 + e(3.938−0.314x) ), x, 0), etc.
P Hl
(a)
V0 = 875, L = 8, t = 5,
2 5
V (5) = 875 1 −
= 207.64.
8
The refrigerator will be worth \$207.64
after 5 years.
875 − 207.64
5
133.47
875
(b)
0
V (t) = V0
=
\$133.47 and
=
0.15 or 15%
2
1−
L
t
ln
2
1−
L
=
The percentage rate of change is
2 t
2
V
1
−
ln
1
−
0
V 0 (t)
L
L
100
= 100
V (t)
2 t
V0 1 −
L
2
= 100 ln 1 −
L
year
1790
1800
1830
1860
1880
1900
1920
1940
1960
1980
1990
2000
t
0
1
4
7
9
11
13
15
17
19
20
21
P (t)
3, 867, 087
5, 256, 550
12, 956, 719
30, 207, 500
50, 071, 364
77, 142, 427
108, 425, 601
138, 370, 607
162, 289, 822
178, 782, 499
184, 566, 652
189, 034, 385
(b) This model predicts that the population
will be increasing most rapidly when
t = 12.5 or in 1915.
140
CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
intersects
(c) Writing exercise —
Answers will vary.
37.
x
1
2
x
1
=
3
x
1
=
5
y
=
2−x =
y
=
3−x
y
=
5−x
(0.5)−x = 2x
x
1
The graphs of y = bx and y =
are
b
reflections of each other in the y–axis
(0 < b < 1).
The larger b the steeper the curve.
y
38.
y
y
y
√
y = 4 − ln x
at (1.2373, 3.8935) according to the graphing
utility.
=
40.
can be rewritten as
1
1
1
ln(x + 5) −
ln x =
ln(x2 + 2x)2
ln 5
ln 2
ln 10
Hint: Use the equation writer on the HP48G to
enter the equation (and make a copy on the
stack) and store in the EQ field (use ’EQ’ and
STOre).
According to the SOLVE capability of the
graphing utility.
Use GRAPH, y(x) =, and ZOOM/TRACE on
the TI-85.
√
=
3x = 3x/2
√
=
3−x = 3−x/2
= 3−x = 3−x
x = 1.06566543483
is a solution of this equation. Plot
f (x) = log5 (x + 5) − log2 x − 2 log10 (x2 + 2x)
The graphs of y = 3bx and y = 3−bx are
reflections of each other in the y–axis (0 < b).
The larger b the steeper the curve.
because the message “sign reversal” (on the
HP48G) indicates another root.
x < 0 leads to a complex solution (as it should
since the argument of a logarithm must be
positive).
Plotting on 0 < x < 500, 000 did not reveal
another root.
There is no other root because x2 increases
much more rapidly than any other argument,
making f (x) monotonically decreasing.
41.
39. Hint: Use the equation writer on the HP48G to
enter the function or y(x) = on the TI-85.
y = 3x
log5 (x + 5) − log2 x = 2 log10 (x2 + 2x)
= f (x) = ln(1 + x2 ) and
1
y = g(x) =
x
intersect at (1.166, 0.858) according to the
graphing utility.
y
4.5. REVIEW PROBLEMS
141
even when n is large.
42.
√
√
√ √
n ( n) n+1 ( n + 1) n
8
22.63
22.36
9
32.27
31.62
12
88.21
85.00
20
957.27
904.84
25
3, 665
3, 447
31
16, 528
15, 494
37
68, 159
63, 786
38
85, 679
80, 166
43
261, 578
244, 579
50 1, 165, 565
1, 089, 362
100 1.12 × 1010 1.05 × 1010
1, 000 2.87 × 1047 2.76 × 1047
√
Thus (n + 1)
n
≤n
√
n+1
Mathematically one could reason as follows:
√
(x + 1) x
√
lim
x→∞ x x+1
√x + 1
x+1
≤ lim
x→∞
x
√x + 1
x+1
= lim exp ln
x→∞
x

 x+1
 ln

x

= lim exp 
 (x + 1)−1/2 
x→∞
x(−1/x2 )
x→∞ (x + 1)(−1/2)(x + 1)−3/2
√
2 x+1
= exp lim
x→∞
x
2
= e0 = 1
= exp lim √
x→∞ 2 x + 1
=
exp
lim
√
Thus (n + 1)
n
≤n
√
n+1
```