solutions to chapter review questions, chapter 3park/fall2013/precalculus/ch3_sol.pdf ·...

Instructor’s Solutions Manual, Chapter 3 Review Question 1

Solutions to Chapter Review Questions, Chapter 3

1. Explain why√

52 = 5.

solution The square root of 5, denoted√

5, is defined as the positivenumber whose square equals 5. Thus

√5

2 = 5.


2. Give an example of a number t such that√t2 �= t.

solution Any negative number will work. To give a specific example,let t = −3. Then t2 = 9 and hence

√t2 = √9 = 3. Thus for t = −3, we

have√t2 �= t.


3. Show that (29+ 12√

5)1/2 = 3+ 2√

5.

solution We need to show that the square of 3+ 2√

5 equals29+ 12

√5. Here is that computation:

(3+ 2√

5)2 = 9+ 2 · 3 · 2√

5+ 4 · 5 = 29+ 12√

5.


4. Evaluate 327/5.

solution327/5 = (321/5)7 = 27 = 128


5. Expand (4− 3√

5x)2.

solution

(4− 3√

5x)2 = 16− 2 · 4 · 3√

5x + 9 · 5x = 16− 24√

5x + 45x


6. What is the domain of the function f defined by f(x) = x3/5?

solution Recall that x3/5 is defined to equal (x1/5)3 whenever thismakes sense. Because 5 is an odd number, x1/5 makes sense for everyreal number x. Thus x3/5 is defined for every real number x. Hence thedomain of f is the set of real numbers.


7. What is the domain of the function f defined by f(x) = (x − 5)3/4?

solution Recall that t3/4 is defined to equal (t1/4)3 whenever thismakes sense. Because 4 is an even number, t1/4 makes sense only whent ≥ 0. Thus t3/4 is defined only when t ≥ 0, which implies that(x − 5)3/4 is defined only when x − 5 ≥ 0, which is equivalent to theinequality x ≥ 5. Hence the domain of f is the interval [5,∞).


8. Find the inverse of the function f defined by

f(x) = 3+ 2x4/5.

solution To find the inverse function of f , let y be a number. Weneed to solve the equation

3+ 2x4/5 = y

for x. To do this, subtract 3 from both sides and then divide by 2,getting

x4/5 = y − 32

.

Now raise both sides to the power 54 , getting

x =(y − 3

2

)5/4.

Thus

f−1(y) =(y − 3

2

)5/4.

The number insider the parentheses above must be nonnegative so thatit can be raised to the power 5

4 . In other words, we must have y ≥ 3.Thus the domain of f−1 is the interval [3,∞).


9. Find a formula for (f ◦ g)(x), where

f(x) = 3x√

32 and g(x) = x√

2.

solution

(f ◦ g)(x) = f (g(x)) = f (x√2) = 3(x√

2)√32 = 3x√

2·√32 = 3x√

64 = 3x8


10. Explain how logarithms are defined.

solution If b is a positive number with b �= 1 and y is a positivenumber, then logb y is defined to be the number x such that bx = y .


11. What is the domain of the function f defined by f(x) = log2 x?

solution Logarithms are defined only for positive numbers. Thus thedomain of f is the interval (0,∞).


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 rangeof f is the set of real numbers.


13. Explain why3log3 7 = 7.

solution Let x = log3 7. By the definition of logarithm, we have3x = 7. In other words, 3log3 7 = 7.


14. Explain whylog5 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.


15. Without using a calculator or computer, estimate the number of digitsin 21000.

solution Note that

21000 = 210×100 = (210)100 = 1024100 ≈ (103)100 = 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 301digits.

[Actually 21000 has 302 digits, so our estimate is close but not exact.]


16. Find all numbers x such that

logx + log(x + 2) = 1.

solution We have

1 = logx + log(x + 2) = log(x(x + 2)

).

Thusx(x + 2) = 10,

which can be written as

x2 + 2x − 10 = 0.

Using the quadratic formula, we get

x = −2±√4+ 4 · 102

= −2±√4 · 112

= −1±√11.

Thus the only possibilities for x are −1−√11 and −1+√11. However,taking x to be −1−√11 makes no sense in the expressionlogx + log(x + 2) because the logarithm of a negative number is notdefined. Thus the only number x satisfying the equationlogx + log(x + 2) = 1 is x = −1+√11.


17. Evaluate log5

√125.

solution

log5

√125 = log5

√53 = log5 (53)1/2 = log5 53/2 = 3

2


18. Find a number b such that logb 9 = −2.

solution The equation logb 9 = −2 means that

9 = b−2 = 1b2.

Thus

b2 = 19,

which implies that b = 13 (the possibility that b = −1

3 is excludedbecause the base for a logarithm must be a positive number).


19. How many digits does 47000 have?

solution To find the number of digits in 47000, we first compute itlogarithm:

log 47000 = 7000 log 4 ≈ 4214.4.

Thus 47000 has 4215 digits.


20. At the time this book was written, the largest known prime numbernot of the form 2n − 1 was 19249 · 213018586 + 1. How many digits doesthis prime number have?

solution The decimal representation of the number19249 · 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 thenumber 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 hence19249 · 213018586 + 1 also has 3918990 digits.


21. Find the smallest integer m such that

8m > 10500.

solution Taking logarithms, we see that the inequality 8m > 10500 isequivalent to

log 8m > log 10500,

which can be rewritten as

m log 8 > 500,

which is equivalent to the inequality

m >500log 8

≈ 553.7.

The smallest integer that is greater than 553.7 is 554. Thus 554 is thesmallest integer m such that 8m > 10500.


22. Find the largest integer k such that

15k < 11900.

solution Taking logarithms, we see that the inequality 15k < 11900 isequivalent 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

log 15≈ 796.9.

The largest integer that is less than 796.9 is 796. Thus 796 is thelargest integer k such that 15k < 11900.


23. Which of the expressions

logx + logy and (logx)(logy)

can be rewritten using only one log?

solution The expression logx + logy can be rewritten as log(xy).

There is no nice way to rewrite (logx)(logy).


24. Which of the expressions

logx − logy andlogxlogy

can be rewritten using only one log?

solution The expression logx − logy can be rewritten as log xy .

There is no nice way to rewrite logxlogy .


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. Weneed 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

log3(7x + 2) = y − 45

.

Thus7x + 2 = 3(y−4)/5,

which implies that

x = 3(y−4)/5 − 27

.

Thus

f−1(y) = 3(y−4)/5 − 27

.



f(x) = 74x and g(x) = log7 x.

solution

(f ◦ g)(x) = f (g(x)) = f(log7 x) = 74 log7 x = (7log7 x)4 = x4



f(x) = log2 x and g(x) = 25x−9.

solution

(f ◦ g)(x) = f (g(x)) = f(25x−9) = log2 25x−9 = 5x − 9


28. Evaluate log3.2 456.

solution Most calculators can evaluate logarithms only with base 10or base e (to be discussed in the next chapter). Thus we use the changeof base formula to convert this logarithm with base 3.2 into a logarithmwith base 10:

log3.2 456 = log 456log 3.2

≈ 5.26371.


29. Suppose log6 t = 4.3. Evaluate log6 t200.

solutionlog6 t200 = 200 log6 t = 200× 4.3 = 860


30. Suppose log7w = 3.1 and log7 z = 2.2. Evaluate

log749w2

z3.

solution

log749w2

z3 = log7(49w2)− log7 z3

= log7 49+ log7w2 − 3 log7 z

= 2+ 2 log7w − 3 · 2.2

= −4.6+ 2 · 3.1

= 1.6


31. Suppose $7000 is deposited in a bank account paying 4% interest peryear, compounded 12 times per year. How much will be in the bankaccount at the end of 50 years?

solution After 50 years, the amount in this bank account will be

7000(1+ 0.04

12

)12×50 ≈ 51551.6

dollars.


32. Suppose $5000 is deposited in a bank account that compoundsinterest four times per year. The bank account contains $9900 after 13years. What is the annual interest rate for this bank account?

solution Let r denote the annual interest rate for this bank account.Then

5000(1+ r4 )

4×13 = 9900.

The equation above can be written as

(1+ r4 )

52 = 9950 .

Thus1+ r

4 = (9950)

1/52.

Solving for r , we get

r = 4(9950)

1/52 − 4 ≈ 0.05289.

Thus the annual interest rate is approximately 5.289%.


33. A colony that initially contains 100 bacteria cells is growingexponentially, doubling in size every 75 minutes. Approximately howmany bacteria cells will the colony have after 6 hours?

solution The statement of this question involves both minutes andhours. As usual, calculations should use the same unit of time. Thuswe 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), thecolony will contain

100× 2360/75 ≈ 2786

bacteria cells.


34. A colony of bacteria is growing exponentially, doubling in size every50 minutes. How many minutes will it take for the colony to become sixtimes its current size?

solution Because the colony of bacteria is doubling in size every 50minutes, after t minutes the colony has increased in size by a factor of2t/50. For the colony to become six times its current size, we need tofind 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 = t50 log 2.

Thus

t = 50 log 6log 2

≈ 129.25.

Thus the colony will become six times its current size after slightlymore than 129 minutes.


35. A colony of bacteria is growing exponentially, increasing in size from200 to 500 cells in 100 minutes. How many minutes does it take thecolony to double in size?

solution Suppose the colony doubles in size every d minutes. Then

200× 2100/d = 500.

Thus2100/d = 5

2 .

Taking logarithms of both sides gives

100d log 2 = log 5

2 .

Thus

d = 100 log 2

log 52

≈ 75.6.

Thus this colony of bacteria doubles in size approximately every 75.6minutes.


36. Explain why a population cannot have exponential growth indefinitely.

solution Exponential growth of a population cannot continueindefinitely because eventually exponential growth leads to populationnumbers that are too large to be physically possible. For example, thereare on the order of 1080 atoms in the universe (according to currentestimates). If exponential growth continued indefinitely, theneventually the population would be larger than the number of atoms inthe universe, which is clearly impossible. Actually exponential growthof a population would stop considerably before reaching the number ofatoms in the universe, because food and other resources would run outmuch earlier.


37. About how many years will it take for a sample of cesium-137, whichhas a half-life of 30 years, to have only 3% as much cesium-137 as theoriginal sample?

solution Because cesium-137 has a half-life of 30 years, after t yearsonly 2−t/30 times the amount in the original sample of cesium-137 willremain. Thus we seek a number t such that

0.03 = 2−t/30.

Taking logarithms of both sides, we get

log 0.03 = log 2−t/30 = − t30 log 2.

Thus

t = −30 log 0.03log 2

≈ 151.8.

In other words, a sample of cesium-137 will contain only 3% as muchcesium-137 as the original sample after about 152 years.


38. How many more times intense is an earthquake with Richtermagnitude 6.8 than an earthquake with Richter magnitude 6.1?

solution Each increase of 1 in Richter magnitude multiplies theintensity by a factor of 10. Thus an increase of 0.7 (which is thedifference between 6.8 and 6.1) increases the intensity by a factor of100.7. Because 100.7 ≈ 5, an earthquake with Richter magnitude 6.8 isapproximately five times more intense than an earthquake with Richtermagnitude 6.1


39. Explain why adding ten decibels to a sound multiplies the intensity ofthe sound by a factor of 10.

solution A sound with intensity E has d decibels, where

d = 10 logEE0

;

here E0 is the intensity of a sound at the threshold of human hearing.Adding 10 to both sides of the equation above gives

d+ 10 = 10+ 10 logEE0

= 10(1+ log

EE0

)= 10

(log 10+ log

EE0

)= 10 log

10EE0

.

The equation above shows that the intensity of a sound with d+ 10decibels is 10E. In other words, adding ten decibels to a soundmultiplies the intensity of the sound by a factor of 10.


40. Most stars have an apparent magnitude that is a positive number.However, four stars (not counting the sun) have an apparent magnitudethat is a negative number. Explain how a star can have a negativemagnitude.

solution A star with brightness b has apparent magnitude

52

logb0

b,

where b0 is approximately the brightness of the star Vega. However,four stars (not counting our sun) are brighter than Vega; for those fourstars the ratio b0

b is less than 1. A number less than 1 has a negativelogarithm, and thus those four stars have a negative apparentmagnitude.

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