Instructor’s Solutions Manual, Chapter 3
Review Question 1
Solutions to Chapter Review Questions, Chapter 3
1. Explain why
√ 2
5 = 5.
√
solution The square root of 5, denoted 5, is defined as the positive
√ 2
number whose square equals 5. Thus 5 = 5.
Instructor’s Solutions Manual, Chapter 3
2. Give an example of a number t such that
Review Question 2
√
t 2 = t.
solution Any negative number will work. To give a specific example,
√
√
let t = −3. Then t 2 = 9 and hence t 2 = 9 = 3. Thus for t = −3, we
√
have t 2 = t.
Instructor’s Solutions Manual, Chapter 3
Review Question 3
√
√
3. Show that (29 + 12 5)1/2 = 3 + 2 5.
√
solution We need to show that the square of 3 + 2 5 equals
√
29 + 12 5. Here is that computation:
√
√
√
(3 + 2 5)2 = 9 + 2 · 3 · 2 5 + 4 · 5 = 29 + 12 5.
Instructor’s Solutions Manual, Chapter 3
Review Question 4
4. Evaluate 327/5 .
solution
7
327/5 = (321/5 ) = 27 = 128
Instructor’s Solutions Manual, Chapter 3
Review Question 5
√
5. Expand (4 − 3 5x)2 .
solution
√
√
√
(4 − 3 5x)2 = 16 − 2 · 4 · 3 5x + 9 · 5x = 16 − 24 5x + 45x
Instructor’s Solutions Manual, Chapter 3
Review Question 6
6. What is the domain of the function f defined by f (x) = x 3/5 ?
3
solution Recall that x 3/5 is defined to equal (x 1/5 ) whenever this
makes sense. Because 5 is an odd number, x 1/5 makes sense for every
real number x. Thus x 3/5 is defined for every real number x. Hence the
domain of f is the set of real numbers.
Instructor’s Solutions Manual, Chapter 3
Review Question 7
7. What is the domain of the function f defined by f (x) = (x − 5)3/4 ?
3
solution Recall that t 3/4 is defined to equal (t 1/4 ) whenever this
makes sense. Because 4 is an even number, t 1/4 makes sense only when
t ≥ 0. Thus t 3/4 is defined only when t ≥ 0, which implies that
(x − 5)3/4 is defined only when x − 5 ≥ 0, which is equivalent to the
inequality x ≥ 5. Hence the domain of f is the interval [5, ∞).
Instructor’s Solutions Manual, Chapter 3
Review Question 8
8. Find the inverse of the function f defined by
f (x) = 3 + 2x 4/5 .
solution To find the inverse function of f , let y be a number. We
need to solve the equation
3 + 2x 4/5 = y
for x. To do this, subtract 3 from both sides and then divide by 2,
getting
y −3
.
x 4/5 =
2
Now raise both sides to the power
x=
5
4,
getting
y − 3 5/4
.
2
Thus
f −1 (y) =
y − 3 5/4
.
2
The number insider the parentheses above must be nonnegative so that
it can be raised to the power 54 . In other words, we must have y ≥ 3.
Thus the domain of f −1 is the interval [3, ∞).
Instructor’s Solutions Manual, Chapter 3
Review Question 9
9. Find a formula for (f ◦ g)(x), where
f (x) = 3x
√
32
and g(x) = x
√
2
.
solution
√ √
√
√ √32
√ = 3x 2· 32 = 3x 64 = 3x 8
(f ◦ g)(x) = f g(x) = f x 2 = 3 x 2
Instructor’s Solutions Manual, Chapter 3
Review Question 10
10. Explain how logarithms are defined.
solution If b is a positive number with b = 1 and y is a positive
number, then logb y is defined to be the number x such that bx = y.
Instructor’s Solutions Manual, Chapter 3
Review Question 11
11. What is the domain of the function f defined by f (x) = log2 x?
solution Logarithms are defined only for positive numbers. Thus the
domain of f is the interval (0, ∞).
Instructor’s Solutions Manual, Chapter 3
Review Question 12
12. What is the range of the function f defined by f (x) = log2 x?
solution If t is any real number, then
log2 2t = t.
Thus every real number t is in the range of f . In other words, the range
of f is the set of real numbers.
Instructor’s Solutions Manual, Chapter 3
Review Question 13
13. Explain why
3log3 7 = 7.
solution Let x = log3 7. By the definition of logarithm, we have
3x = 7. In other words, 3log3 7 = 7.
Instructor’s Solutions Manual, Chapter 3
Review Question 14
14. Explain why
log5 5444 = 444.
solution By definition of the logarithm, if log5 5444 = x then
5x = 5444 .
The equation above shows that x = 444. Thus log5 5444 = 444.
Instructor’s Solutions Manual, Chapter 3
Review Question 15
15. Without using a calculator or computer, estimate the number of digits
in 21000 .
solution Note that
100
21000 = 210×100 = (210 )
100
= 1024100 ≈ (103 )
= 10300 .
Because the decimal representation of 10300 is 1 followed by 300 0’s,
we know that 10300 has 301 digits. Thus we estimate that 21000 has 301
digits.
[Actually 21000 has 302 digits, so our estimate is close but not exact.]
Instructor’s Solutions Manual, Chapter 3
Review Question 16
16. Find all numbers x such that
log x + log(x + 2) = 1.
solution We have
1 = log x + log(x + 2) = log x(x + 2) .
Thus
x(x + 2) = 10,
which can be written as
x 2 + 2x − 10 = 0.
Using the quadratic formula, we get
√
√
4 + 4 · 10
−2 ± 4 · 11
=
= −1 ± 11.
x=
2
2
√
√
Thus the only possibilities for x are −1 − 11 and −1 + 11. However,
√
taking x to be −1 − 11 makes no sense in the expression
log x + log(x + 2) because the logarithm of a negative number is not
defined. Thus the only number x satisfying the equation
√
log x + log(x + 2) = 1 is x = −1 + 11.
−2 ±
√
Instructor’s Solutions Manual, Chapter 3
17. Evaluate log5
Review Question 17
√
125.
solution
log5
√
1/2
125 = log5 53 = log5 (53 )
= log5 53/2 =
3
2
Instructor’s Solutions Manual, Chapter 3
Review Question 18
18. Find a number b such that logb 9 = −2.
solution The equation logb 9 = −2 means that
9 = b−2 =
Thus
b2 =
1
1
.
b2
1
,
9
1
which implies that b = 3 (the possibility that b = − 3 is excluded
because the base for a logarithm must be a positive number).
Instructor’s Solutions Manual, Chapter 3
19.
Review Question 19
How many digits does 47000 have?
solution To find the number of digits in 47000 , we first compute it
logarithm:
log 47000 = 7000 log 4 ≈ 4214.4.
Thus 47000 has 4215 digits.
Instructor’s Solutions Manual, Chapter 3
20.
Review Question 20
At the time this book was written, the largest known prime number
not of the form 2n − 1 was 19249 · 213018586 + 1. How many digits does
this prime number have?
solution The decimal representation of the number
19249 · 213018586 + 1 is not a power of 10 because it is a prime number.
Thus the number of digits in 19249 · 213018586 + 1 is the same as the
number of digits in 19249 · 213018586 .
To find the number of digits in 19249 · 213018586 , we take its logarithm:
log(19249 · 213018586 ) = log 19249 + log 213018586
= log 19249 + 13018586 log 2
≈ 3918989.2.
Thus 19249 · 213018586 has 3918990 digits, and hence
19249 · 213018586 + 1 also has 3918990 digits.
Instructor’s Solutions Manual, Chapter 3
21.
Review Question 21
Find the smallest integer m such that
8m > 10500 .
solution Taking logarithms, we see that the inequality 8m > 10500 is
equivalent to
log 8m > log 10500 ,
which can be rewritten as
m log 8 > 500,
which is equivalent to the inequality
m>
500
≈ 553.7.
log 8
The smallest integer that is greater than 553.7 is 554. Thus 554 is the
smallest integer m such that 8m > 10500 .
Instructor’s Solutions Manual, Chapter 3
22.
Review Question 22
Find the largest integer k such that
15k < 11900 .
solution Taking logarithms, we see that the inequality 15k < 11900 is
equivalent to
log 15k < log 11900 ,
which can be rewritten as
k log 15 < 900 log 11,
which is equivalent to the inequality
k<
900 log 11
≈ 796.9.
log 15
The largest integer that is less than 796.9 is 796. Thus 796 is the
largest integer k such that 15k < 11900 .
Instructor’s Solutions Manual, Chapter 3
Review Question 23
23. Which of the expressions
log x + log y
and
(log x)(log y)
can be rewritten using only one log?
solution The expression log x + log y can be rewritten as log(xy).
There is no nice way to rewrite (log x)(log y).
Instructor’s Solutions Manual, Chapter 3
Review Question 24
24. Which of the expressions
log x − log y
and
log x
log y
can be rewritten using only one log?
solution The expression log x − log y can be rewritten as log
There is no nice way to rewrite
log x
log y .
x
y.
Instructor’s Solutions Manual, Chapter 3
Review Question 25
25. Find a formula for the inverse of the function f defined by
f (x) = 4 + 5 log3 (7x + 2).
solution To find the inverse function of f , let y be a number. We
need to solve the equation
4 + 5 log3 (7x + 2) = y
for x. To do this, subtract 4 from both sides and then divide by 5,
getting
y −4
.
log3 (7x + 2) =
5
Thus
7x + 2 = 3(y−4)/5 ,
which implies that
x=
3(y−4)/5 − 2
.
7
Thus
f −1 (y) =
3(y−4)/5 − 2
.
7
Instructor’s Solutions Manual, Chapter 3
Review Question 26
26. Find a formula for (f ◦ g)(x), where
f (x) = 74x
and g(x) = log7 x.
solution
4
(f ◦ g)(x) = f g(x) = f (log7 x) = 74 log7 x = (7log7 x ) = x 4
Instructor’s Solutions Manual, Chapter 3
Review Question 27
27. Find a formula for (f ◦ g)(x), where
f (x) = log2 x
and
g(x) = 25x−9 .
solution
(f ◦ g)(x) = f g(x) = f (25x−9 ) = log2 25x−9 = 5x − 9
Instructor’s Solutions Manual, Chapter 3
28.
Review Question 28
Evaluate log3.2 456.
solution Most calculators can evaluate logarithms only with base 10
or base e (to be discussed in the next chapter). Thus we use the change
of base formula to convert this logarithm with base 3.2 into a logarithm
with base 10:
log 456
≈ 5.26371.
log3.2 456 =
log 3.2
Instructor’s Solutions Manual, Chapter 3
Review Question 29
29. Suppose log6 t = 4.3. Evaluate log6 t 200 .
solution
log6 t 200 = 200 log6 t = 200 × 4.3 = 860
Instructor’s Solutions Manual, Chapter 3
Review Question 30
30. Suppose log7 w = 3.1 and log7 z = 2.2. Evaluate
log7
49w 2
.
z3
solution
log7
49w 2
z3
= log7 (49w 2 ) − log7 z3
= log7 49 + log7 w 2 − 3 log7 z
= 2 + 2 log7 w − 3 · 2.2
= −4.6 + 2 · 3.1
= 1.6
Instructor’s Solutions Manual, Chapter 3
31.
Review Question 31
Suppose $7000 is deposited in a bank account paying 4% interest per
year, compounded 12 times per year. How much will be in the bank
account at the end of 50 years?
solution After 50 years, the amount in this bank account will be
7000 1 +
dollars.
0.04 12×50
12
≈ 51551.6
Instructor’s Solutions Manual, Chapter 3
32.
Review Question 32
Suppose $5000 is deposited in a bank account that compounds
interest four times per year. The bank account contains $9900 after 13
years. What is the annual interest rate for this bank account?
solution Let r denote the annual interest rate for this bank account.
Then
r
5000(1 + 4 )4×13 = 9900.
The equation above can be written as
r
(1 + 4 )52 =
99
50 .
Thus
1+
r
4
99
= ( 50 )1/52 .
Solving for r , we get
99
r = 4( 50 )1/52 − 4 ≈ 0.05289.
Thus the annual interest rate is approximately 5.289%.
Instructor’s Solutions Manual, Chapter 3
33.
Review Question 33
A colony that initially contains 100 bacteria cells is growing
exponentially, doubling in size every 75 minutes. Approximately how
many bacteria cells will the colony have after 6 hours?
solution The statement of this question involves both minutes and
hours. As usual, calculations should use the same unit of time. Thus
we convert 6 hours to 360 minutes.
After t minutes, the colony will contain
100 × 2t/75
bacteria cells. Thus after 360 minutes (which equals 6 hours), the
colony will contain
100 × 2360/75 ≈ 2786
bacteria cells.
Instructor’s Solutions Manual, Chapter 3
34.
Review Question 34
A colony of bacteria is growing exponentially, doubling in size every
50 minutes. How many minutes will it take for the colony to become six
times its current size?
solution Because the colony of bacteria is doubling in size every 50
minutes, after t minutes the colony has increased in size by a factor of
2t/50 . For the colony to become six times its current size, we need to
find a number t such that
6 = 2t/50 .
To solve the equation above for t, take logarithms of both sides, getting
log 6 = log 2t/50 =
Thus
t=
t
50
log 2.
50 log 6
≈ 129.25.
log 2
Thus the colony will become six times its current size after slightly
more than 129 minutes.
Instructor’s Solutions Manual, Chapter 3
35.
Review Question 35
A colony of bacteria is growing exponentially, increasing in size from
200 to 500 cells in 100 minutes. How many minutes does it take the
colony to double in size?
solution Suppose the colony doubles in size every d minutes. Then
200 × 2100/d = 500.
Thus
5
2100/d = 2 .
Taking logarithms of both sides gives
100
d
Thus
d=
log 2 = log 52 .
100 log 2
log
5
2
≈ 75.6.
Thus this colony of bacteria doubles in size approximately every 75.6
minutes.
Instructor’s Solutions Manual, Chapter 3
Review Question 36
36. Explain why a population cannot have exponential growth indefinitely.
solution Exponential growth of a population cannot continue
indefinitely because eventually exponential growth leads to population
numbers that are too large to be physically possible. For example, there
are on the order of 1080 atoms in the universe (according to current
estimates). If exponential growth continued indefinitely, then
eventually the population would be larger than the number of atoms in
the universe, which is clearly impossible. Actually exponential growth
of a population would stop considerably before reaching the number of
atoms in the universe, because food and other resources would run out
much earlier.
Instructor’s Solutions Manual, Chapter 3
37.
Review Question 37
About how many years will it take for a sample of cesium-137, which
has a half-life of 30 years, to have only 3% as much cesium-137 as the
original sample?
solution Because cesium-137 has a half-life of 30 years, after t years
only 2−t/30 times the amount in the original sample of cesium-137 will
remain. Thus we seek a number t such that
0.03 = 2−t/30 .
Taking logarithms of both sides, we get
t
log 0.03 = log 2−t/30 = − 30 log 2.
Thus
t=−
30 log 0.03
≈ 151.8.
log 2
In other words, a sample of cesium-137 will contain only 3% as much
cesium-137 as the original sample after about 152 years.
Instructor’s Solutions Manual, Chapter 3
38.
Review Question 38
How many more times intense is an earthquake with Richter
magnitude 6.8 than an earthquake with Richter magnitude 6.1?
solution Each increase of 1 in Richter magnitude multiplies the
intensity by a factor of 10. Thus an increase of 0.7 (which is the
difference between 6.8 and 6.1) increases the intensity by a factor of
100.7 . Because 100.7 ≈ 5, an earthquake with Richter magnitude 6.8 is
approximately five times more intense than an earthquake with Richter
magnitude 6.1
Instructor’s Solutions Manual, Chapter 3
Review Question 39
39. Explain why adding ten decibels to a sound multiplies the intensity of
the sound by a factor of 10.
solution A sound with intensity E has d decibels, where
d = 10 log
E
;
E0
here E0 is the intensity of a sound at the threshold of human hearing.
Adding 10 to both sides of the equation above gives
E
E0
E
d + 10 = 10 + 10 log
= 10 1 + log
E0
= 10 log 10 + log
= 10 log
E
E0
10E
.
E0
The equation above shows that the intensity of a sound with d + 10
decibels is 10E. In other words, adding ten decibels to a sound
multiplies the intensity of the sound by a factor of 10.
Instructor’s Solutions Manual, Chapter 3
Review Question 40
40. Most stars have an apparent magnitude that is a positive number.
However, four stars (not counting the sun) have an apparent magnitude
that is a negative number. Explain how a star can have a negative
magnitude.
solution A star with brightness b has apparent magnitude
b0
5
log
,
2
b
where b0 is approximately the brightness of the star Vega. However,
four stars (not counting our sun) are brighter than Vega; for those four
b
stars the ratio b0 is less than 1. A number less than 1 has a negative
logarithm, and thus those four stars have a negative apparent
magnitude.