000200010271835410 CH08 p000-000 · There are many possible solutions. For example, ... CME_Int_L3_CSM_CH08 03/03/2014 14:43 Page 254 (a perfect square) or 8 because 72 ÷8 = 9 (a

Chapter Exponential and Logarithmic Functions pages 2–.....................................................................

INVESTIGATION 8A PROPERTIES OF

. Getting Started

For You to Explore

1.

Input Output �

0 1 31 4 32 7 33 10 34 13 35 16 3

2.

Input Output �

0 1 01 1 22 3 43 7 64 13 85 21 10

3.

Input Output �

0 1 21 3 62 9 183 27 544 81 1625 243 486

4.

Input Output �

0 3 121 15 602 75 3003 375 15004 1875 75005 9375 37,500

5.

Input Output �

0 1 − 12

1 12 − 1

4

2 14 − 1

8

3 18 − 1

16

4 116 − 1

32

5 132 − 1

64

6. In the last problem, f (x) = (12

)x = 2−x , the outputs arethe negatives of the inputs. So, a good guess might beg(x) = 2x . The table for g(x) is:

Input Output �

0 1 11 2 22 4 43 8 84 16 165 32 32

The output column equals the difference column.g(x) = 0 also works.

7. (a)

Input Output ÷0 1 41 4 1.752 7 1.433 10 1.34 13 1.235 16 1.19

(b) The numbers in the ratio column appear to beapproaching 1. This makes sense because as theoutputs get larger and larger, adding three to theprevious output doesn’t change it as much as whenthe outputs are small.

8. (a) b(x) = x2 − x + 1

Input Output ÷0 1 11 1 32 3 2.333 7 1.864 13 1.625 21 1.48

EXPONENTIAL FUNCTIONS

8 01

825 288

Mathematics III Solution Manual • Chapter , page 2s 8 52

CME_Int_L3_CSM_CH08 03/03/2014 14:43 Page 252

(b) c(x) = 3x

Input Output ÷0 1 31 3 32 9 33 27 34 81 35 243 3

(c) d(x) = 3 · 5x

Input Output ÷0 3 51 15 52 75 53 375 54 1875 55 9375 5

(d) f (x) = (12

)xInput Output ÷

0 1 12

1 12

12

2 14

12

3 18

12

4 116

12

5 132

12

9. (a) If a table has constant differences, the function islinear and the rule will be in the form f (x) =(constant difference) · x + (a constant). To find theconstant, look at the output for an input of 0. So, thefunction will be f (x) = ( constant difference) · x+(f (0)).

(b) If the table has constant ratios, then the function isexponential and the rule will be in the form

= ( a constant) ( constant ratio)x . To find theconstant, look at the output for an input of 0. So, thefunction will be f (x) = f (0) · ( common ratio)x .

10. If the common ratio is going to equal the output, yourtable will look like this:

Input Output ÷0 f (0) f (0)

1 f (1) f (1)

2 f (2) f (2)

3 f (3) f (3)

4 f (4) f (4)

5 f (5) f (5)

If the common ratios = the outputs, f (2)

f (1)= f (1),

f (3)

f (2)= f (2), and f (4)

f (3)= f (3). So, f (2) = (f (1))2,

f (3) = (f (2))2, and f (4) = (f (3))2. Now your tablelooks like this:

Input Output ÷0 f (0) f (0)

1 f (0)2 f (0)2

2 f (1)2 = f (0)4 f (0)4

3 f (2)2 = f (0)8 f (0)8

4 f (3)2 = f (0)16 f (0)16

5 f (0)32 f (0)32

You can let f (0) be some number, say 3, the exponentsfor the function are powers of 2. One possible solution isf (x) = 3(2x )

On Your Own

11. (a) For x2 = 5, take the square root of both sides.Remember that the solution could be the positive ornegative square root. x = ±√

5(b) Estimate the solution using your calculator. Since

22 = 4 and 23 = 8, a good guess to start would be alittle more than 2. x ≈ 2.322. Many students willchoose this equation as the most difficult for them tosolve.

(c) x = 52 = 25(d) Since 8 is a power of 2, rewrite the equation as

23 = 2x and x = 3.12.

O�2 2

2

4

6

8

x

y

13.√

2 ≈ 1.414, and x = 1.414 is between x = 1 and x =Since the graph of y = 3x is continuous, which means(informally) that you can trace it without picking up yourpencil, and increasing, you can say that 3

√2 must be

between 31 and 32, or between 3 and 9.14. One way you could make a more accurate estimate of

3√

2 is by “zooming” in on the graph to find a point whosex-coordinate is close to 1.414. . . You could also use yourcalculator to evaluate 3

√2, either in the home screen or in

the graph screen. See the Technology Appendix for someideas about how to do this.

15. (a) f (a) = 3 · 2a

Input: a Output: f (a)

−2 34

−1 32

0 31 62 12

f (x) ·2.

Mathematics III Solution Manual • Chapter , pages 28 53


30 · 2a

Input: a Output: g(a)

−2 304 = 15

2

−1 302 = 15

0 301 602 120

(c) h(a) = 15 · 5a

Input: a Output: h(a)

0 15

1 12 53 254 125

(d) j (a) = 27 · ( 13

)aInput: a Output: j (a)

−1 810 271 92 33 1

16. (a) k(a) = 4 · 3a

(b) L(h) = 100 · ( 12

)h(c) p(x) = 2 · ( 1

4

)x(d) Q(n) = 8 · ( 3

2

)n17.

x f (x) = 2x

0 122.639

1.41 2.6571.414 2.6651.4142 2.6651.41421 2.6651.414213 2.665

18. (a) For an integer a > 0, 5b = 2a > 1, so b > 0.However, 2a is even and 5b is odd in this case, so theycannot be equal. For an integer a < 0, 5b = 2a < 1,so b < 0. Now 2a = 5b implies that 2−a = 5−b (bytaking reciprocals). Here the exponents are againpositive integers, and you get the same contradictionas in the first case. Therefore, you must have a = 0,so 5b = 20 = 1. Hence b = 0.

(b) Since 4 = 22 and 8 = 23, you can write

4c = 8d

(22)c = (23)d

22c = 23d

2c = 3d

There are many possible solutions. For example,c = 3 and d = 2, or c = 6 and d = 4. You can writec = 3

2 · d . If you choose even integers for d, you willget an integer value for c.

Maintain Your Skills

19. a, b, d, and e can all be found directly. There is nosolution for g. To answer c and f, you can estimate or usea calculator.

(a) f (0) = 20 = 1(b) f

(12

) = 212 = √

2(c) f (π) = 2π ≈ 8.825 (using a calculator)(d) f (f (2)) = f (22) = f (4) = 24 = 16(e)

f (a) = 1

82a = 2−3

a = −3

(f)

f (a) = 7

2a = 7

a ≈ 2.807

This is found using a calculator.(g) 2a will never be a negative number, so there is no

solution.

Check Your Understanding

1. In the first graph, the y values are negative, so a < 0.Therefore, f (x) = −3 · 2x describes this graph. Thesecond graph is decreasing, so 0 < b < 1.f (x) = 3 · ( 1

2

)xdescribes this graph. The third graph

passes through (1, 6). f (x) = 3 · 2x describes this graph.The last graph passes through (1, 15). f (x) = 3 · 5x

describes this graph.2. (a) Given f (x) = a · bx and the points (0, 12) and (2,

you can write the equations 12 = a · b0, so a = 12and 3 = 12 · b2, 1

4 = b2, so b = ± 12 . b � 0, so b =

The function you are looking for is f (x) = 12 · ( 12

(b) Write the equations 12 = a · b2 and 3 = a · b4. Solveeach equation for a, and you get a = 12

b2 and a = 3b

Therefore,

12

b2= 3

b4

12b4 = 3b2

12b4 − 3b2 = 0

3b2(4b2 − 1) = 0

3b2(2b + 1)(2b − 1) = 0

b = 0 or b = ± 12 Since b > 0, b = 1

2 . Substitute to

find a: 3 = a · ( 12

)4, 3 = a · 1

16 , and a = 3 · 16 = 48.The function you are looking for is f (x) = 48 · ( 1

23. (a) To find a possible solution, choose a convenient value

for a. A good choice would be 2 because 72 ÷ 2 =

8.02 Graphs of Exponential Functions

(b) g(a) = ·

11.4

3),

12 .x)

4 .

)x36



(a perfect square) or 8 because 72 ÷ 8 = 9 (a perfectsquare). In the first case, 72 = 2 · b2 and b = ±6,b = −6, b = 6. One equation is f (x) = 2 · 6x . Then,let a = 8 so 72 = 8 · b2 and b = ±3, b = −3, b = 3.Another equation is f (x) = 8 · 3x .

(b) To find the general form, solve the equation72 = a · b2 for a in terms of b, or for b in terms of a.

a = 72b2 or b =

√72a

. The two resulting equations are

f (x) =(

72b2

)· bx and f (x) = a ·

(√72a

)x

.

4. The proof of Lemma 8.2 is precisely the same as that ofLemma 8.1 with all the > signs reversed. Alternately, if0 < b < 1, then b = 1

cwith c > 1. Lemma 8.1 implies

that cx > 1. Now cx = (1b

)x = 1bx , so 1

bx > 1. Therefore,1 > bx . The proof of Theorem 8.2 mimics that ofTheorem . with the signs reversed, or again you canuse b = 1

cwith c > 1. Then f (x) = bx = (

1c

)x = 1cx .

Since the denominator is strictly increasing, the fractionis strictly decreasing.

5. If two graphs are reflections over the y-axis, a point(x, y) on one graph will have an image (−x, y) on theother graph. We need to show that f (−x) = g(x).

f (−x) = 2−x

= 1

2x

=(

1

2

)x

= g(x)

6. x ≈ 2.8077.

2x = 1

7· 4x

7 · 2x = 4x

7 = 4x

2x

7 =(

4

2

)x

7 = 2x

This equation is the same as the one for which youapproximated a solution in Exercise 6, so x ≈ 2.807.

8. (a)

O�8 �4 4 8

8

4

�4

�8

xg

fy

The graphs do not intersect in this window.

(b) The graphs intersect once. The intersection is atroughly x = 35.24 and y = 54.26. You can find thispoint by “zooming out” on your calculator and thenusing the “intersect” function. See the TechnologyAppendix for more about how to do this.

9. (a) 2.66514√

2 ≈ 3.999991207 · · · ≈ 4

(b) Yes,(

2√

2)√

2 = 2√

2·√2 = 22 = 4

10. When you try decimal values to approximate (−2)√

2

with a calculator, you get an error: Non-real result. If youevaluate (−2)

√2, the result is 2

√2 · (−1)

√2. Exponential

functions bx are restricted to b > 0, because (−1)x

cannot be calculated for many fractional values. Forexample, (−1)

12 would be

√−1, which is undefined.

On Your Own

11. The y-intercept is (0, a) becausef (0) = a · b0 = a · 1 = a

12. Test each point to see if it is on the graph.

A. f (0) = −3 · 20 = −3 · 1 = −3 = 1. (0, 1) is NOT onthe graph.

The correct choice is C.13. (a) Next year the cost will increase by 3 percent, so the

price will be 3.99 + .03 · 3.99 or

3.99 · (1 + .03) = 3.99 · 1.03 = $4.11.

The following year, the cost will increase by 3percent again 4.11 · 1.03 = $4.23. Notice that this is

4.11·1.03 = (3.99·1.03) · 1.03 = 3.99 · 1.032 = $4

(b) Multiply $3.99 by 1.03 ten times or by 1.0310.3.99 · 1.0310 = $5.36

(c) Use your calculator. After 55 years the price will be$20.28.

(d) C(n) = 3.99 · 1.03n. The cost will be the originalprice multiplied by (1 + percent increase/100) raisedto the power (number of years)

14. By evaluating 2x for some negative integer values, youcan find that 2−20 = 1

1048576 < 11000000 . Any value less

than −20 will also work.15. Since

√6 ≈ 2.449,

√6 is between 2 and 3. Therefore,

3√

6 must be between 32 = 9 and 33 = 27. You can reasonthis because we see that y = 3x is continuous andincreasing.

16. It is not possible for 2p = 7q if p and q are nonzerointegers. First consider positive integers p and q. If p is apositive integer, 2p is a power of 2, and therefore is aneven integer. If q is a positive integer, 7q is a power of 7,and is an odd integer. Since 2p is even and 7q is odd, theycannot be equal. If p and q are negative integers,

8 1

��

�

B. f (−1) = −3 · 2−1 = −3 · 12 = − 3

2 �= 6. (−1, 6) isNOT on the graph.

C. f (−2) = −3 ·2−2 = −3 · 122 = −3 · 1

4 = − 34 = −.75.

(−2, −.75) IS on the graph.D. f (2) = −3 · 22 = −3 · 4 = −12 �= 36. (2, 36) is

NOT on the graph.

.23



consider 2p = 12−p and 7q = 1

7−q . Following the sameargument, the denominator of the first fraction is alwayseven and the denominator of the second fraction isalways odd, so they are never equal.

17. Since 8 is a multiple of 2, the expressions on each side ofthe equation will both be even numbers, so they could be

equal. In particular, 2pq = 8 = 23, so p

q= 3, and p = 3q.

There are many integers that satisfy this equation.18. (a)

Input Output

−2 14

−1 − 12

0 11 −22 43 −8

(b) The points are alternately positive and negative, soyou cannot connect them with a smooth curvewithout crossing the x-axis. It is not possible for f (x)

to equal zero, so the curve cannot cross the x-axis.19. Suppose first that b and d are positive.

• If b = d , there will be exactly one solution as long as a

and c have the same sign.• If a and c have opposite signs, there is no solution.

Also, if b = d and a = c, there are no solutions.• If b = d and a = c, there will be infinitely many

solutions, because the two sides of the equation will beidentical.Other possibilities arise if you allow any of theparameters to be 0, or if you allow b and d to benegative.

20. (a) By definition, if x is a positive integer, then bx is justb · b · b · · ·︸︷︷︸

x copies

. If b > 0, then b · b · b · b · · · will be the

product of positive numbers and therefore positive. Ifx = 0, then bx = 1 > 0. If x < 0, then bx is thereciprocal of a positive number, so it is positive.

(b) If x is rational, then bx = bpq = q

√bp where p and q

are integers. If b > 0, you just showed that bp > 0and the q

√positive number will be positive.


21. (a) (3√

2)√

2 = 3√

4 = 32 = 9(b) (3

√2)2 = 32·√2

(c) (3−√2)−1 = 3

√2

(d) (3√

8)√

2 = 3√

16 = 34 = 81(e) 3

√2 · 3−√

2 = 3√

2+(−√2) = 30 = 1

(f) (3−√2)−

√2 = 3

√4 = 32 = 9

(g) 3√

2 · 5√

2 = (3 · 5)√

2 = 15√

2

(h) (33√2)

3√2 = 33√4

(i) (33√2)

3√4 = 33√8 = 32 = 9

22.

O 2�2�4

x

yabcdf

4

These graphs are all translations of each other. Forexample, a(x) = 9 when x = 2, but b(x) = 9 whenx = 1. The graph of b(x) is the image of a(x) after atranslation of 1 unit to the left.


1. (a) A(n) = 18 · ( 13

)n(b) B(x) = −2 · 4x

(c) This is not an exponential function, because there isnot a constant ratio between successive terms. 6

4 =(d) D(z) = 1

3 · 6z

2. (a)

A(n) ={

18 if n = 013 · A(n − 1) if n > 0

(b)

B(x) ={

−2 if x = 0

4 · B(x − 1) if x > 0

(c) Not an exponential function.(d)

D(z) ={

13 if z = 0

6 · D(z − 1) if z > 0

3. q(x) = a · bx So you can set up two equations with thegiven points:

100 = a · b3

4 = a · b5

Then divide one equation by the other to get an equationin b alone, and solve for b:

100

4= a · b3

a · b5

25 = b−2

b2 = 1

25

b = 1

5

8.03 Tables of Exponential Functions

�

� � 129



To find a, just substitute your value for b into either ofyour original equations:

100 = a ·(

1

5

)3

100 = a ·(

1

125

)12, 500 = a

The equation is q(x) = 12, 500 · ( 15

)x. If you like, you

can check this equation with your second point.4. b = −1

5 is not a solution because negative bases are notallowed in the definition of an exponential function.Exercise 4 begins by telling you that q(x) is anexponential function.

5. You can find a linear function that solves the problem:f (x) = 2x + 16. Another polynomial function thatsolves the problem is f (x) = 2x + 16 +(x + 3)(x − 2) . . . You can also find an exponentialfunction that solves the problem:

10 = a · b−3

20 = a · b2

Divide:

1

2= b−5

1

2= 1

b5

2 = b5

215 = b

Find a:

10 = a ·(

215

)−3

10 = a · 2− 35

10 · 235 = a

The exponential equation we are looking for is

f (x) = 10 · 235 ·(

215

)x ≈ 15.157 · 1.149x

6. (a) You want to go from 100 to 300 in 5 steps, so b5 =and b = 3

15 . Since T (0) = 100 = a · b0 = a · 1, you

know that a = 100. The function is

T (x) = 100 ·(

315

)x

(b)

x T (x) ÷0 100 3

15 ≈ 1.246

1 100 · 315 ≈ 124.57 3

15 ≈ 1.246

2 100 · 325 ≈ 155.18 3

15 ≈ 1.246

3 100 · 335 ≈ 193.32 3

15 ≈ 1.246

4 100 · 345 ≈ 240.82 3

15 ≈ 1.246

5 300 315 ≈ 1.246

7. (a) Since the car depreciates 20%, it retains 80% of itsvalue. Multiply by 80% or 0.8. After 1 year thevalue is $20,000 · 0.8 = $16,000. After 2 yearsthe value is $16,000 · 0.8 = $12,800. After 3 yearsthe value is $12,800 · 0.8 = $10,240.

(b) This is an exponential function with a = 20,000 andb = 0.8.

V (n) = 20,000 · (0.8)n

(c) Since the function is decreasing, it will eventually beworth less than $1000. Using your calculator, youcan find that V (14) ≈ $880 which is less than $1000.

8.

y1 = a · bx1

y2 = a · bx2

Divide:y1

y2= bx1−x2

(y1

y2

) 1x1−x2 = b

Substitute to find a:

y1 = a ·((

y1

y2

) 1x1−x2

)x1

y1(y1y2

) x1x1−x2

= a

Simplify the expression for a:

a = y

−x2x1−x2

1 · y

x1x1−x2

2

The exponential function that passes through (x1, y1) and2, y2) is

On Your Own

9. Substituting the given values into the function yields

a · b0 = 4

a · b2 = 25

Divide the second equation by the first to get b2 = 254 , so

b = 52 . The first equation simplifies to a = 4. The correct

answer choice is C.10.

n M(n) ÷0 16 1.51 24 1.52 36 1.53 54 1.54 81

3

(x

y = y

−x2x1−x2

1 · y

x1x1−x2

2 ·((

y1

y2

) 1x1−x2

)x



A closed-form rule would be M(n) = 16 · ( 32

)n, since

M(0) = 16 = a and each term is multiplied by 32 = b. To

find a recursive rule, use the fact that you multiply by 1.5to get from one term to the next.

M(n)

{16 if n = 032 · M(n − 1) if n > 0

11. (a)

n F(n) ÷0 1 11 1 22 2 33 6 44 24 55 120 66 720 7

(b) This is not an exponential function because the ratiois not constant.

(c) Since F(1) = 1, F(2) = 2 = 2 · 1, F(3) = 6 =3 · 2 · 1, and F(4) = 24 = 4 · 3 · 2 · 1, you cancalculateF(10) = 10 · 9 · 8 · 7 · · · 3 · 2 · 1 = 3,628,800

12. (a) y = 3 · 2x

(b) y = −514 ·(

514

)x

(c) y = 3 · 213 · ( 1

2

) x6

13. (a) Since the interest is added to Kara’s account, she willhave $1000 + 0.03 · $1000 = $1000 · (1 + 0.03) =$1000 · (1.03) = $1030 in 1 year. Each year youmultiply by 1.03 to get the new amount.$1030 · 1.03 = $1060.90 in 2 years.$1060.90 · 1.03 = $1092.73 in 3 years.

(b) To calculate the amount of money in the account after20 years, you would multiply $1000 by 1.03 twentytimes. $1000 · 1.0320 = $1806.11

14. (a) Multiply by 0.98 each time because if you pay 2%,there will be 100% − 2% = 98% remaining. B(1) =$2000 · 0.98 = $1960; B(2) = $1960 · 0.98 =$1920.80; B(3) = $1920.80 · 0.98 = $1882.38

(b) B(n) = 2000 · (0.98)n

(c) The domain is 0, 1, 2 . . . 12 since payment is madeeach month for 12 months.

15. Use the equation 1000 = 2000 · b12 so b = (12

) 112 ,

b = 0.944. The payment should be100% − 94.4% = 5.6% each month.

16. The functions would not look the same because they havedifferent domains. f (x) is defined for all real numbers,and its graph would be a smooth, connected curve. g(x)

is only defined for nonnegative integers. There would beno points to the left of the y-axis, and points to the rightof the y-axis would only take on integer values of x andwould not be connected to each other.


17. (a) b(5) = 35 = 243(b) b(3) · b(2) = 33 · 32 = 27 · 9 = 243

(c) b(1) = 31 = 3(d) b(3)

b(2)= 33

32 = 33−2 = 31 = 3

(e) b(6) = 36 = 729(f) (b(2))3 = (32)3 = 93 = 729


1. To find C, raise b to the power of x and multiply theresult by a.

2. To find a, raise b to the power of x. Then divide C by thisnumber.

3. To find b, divide C by a and raise the result to thepower 1

x.

4. To find . This gives you Ca

= bx . Inorder to find x, you can approximate the solution bytaking successive guesses at x and use the fact that thefunction bx is increasing if b > 1 and decreasing ifb < 1. (If b = 1 then C

ahad better equal 1, too, or there is

no solution.)5.

x f (x) = 2x

0 11 21.5850 32 42.3219 52.5850 62.8074 73 83.1700 93.3219 10

(a) Since 22 = 4 and 23 = 8, you can fill in those entriesimmediately. To find x if f (x) = 6, write

f (x) = 6

2x = 2 · 3

2x = 21 · 21.5850

2x = 22.5850

x = 2.5850

To find x if f (x) = 9, write

f (x) = 9

2x = 32

2x = (21.5850)2 = 21.5850·2

8.04 Properties of Exponential Functions

x, first divide C by a

18. (a) f (0) = b0 = 1(b) f (−3) = b−3 = 1

b3 = 1f (3)

= 1p

(c) f (8) = b8 = b3 · b5 = f (3) · f (5) = pq

(d) f (6) = b6 = (b3)2 = (f (3))2 = p2

(e) f (15) = b15 = (b3)5 = (f (3))5 = p5 orf (15) = b15 = (b5)3 = (f (5))3 = q3. This lastequation gives you a way to write p in terms of q.



2x = 23.1700

x = 3.1700

To find x if f (x) = 10, write

f (x) = 10

2x = 2 · 5

2x = 21 · 22.3219

2x = 23.3219

x = 3.3219

(b) Since 22.5850 = 6 and 23 = 8, the solution to 2x = 7must be between 2.5850 and 3. By taking successiveguesses at x, you can find x ≈ 2.8074

6. Each of the three rules will change. Let f (x) = a · bx :

(a) f (x + y) = a · bx+y = a · bx · by = f (x) · by Sinceyou need a · by = f (y), multiply by a

a. The result is

f (x + y) = f (x) ·(a

a

)· by = f (x) · a · by

a

= f (x) · f (y)

a

(b) f (x − y) = a · bx−y = a · bx · b−y = f (x) · 1by

Multiply by aa

to get a · by in the denominator. Theresult is

f (x − y) = f (x) ·(a

a

)· 1

by= f (x) · a

a · by= a · f (x)

f (y)

(c) f (xy) = a · bxy = a · (bx)y You need to multiply byay

ay to get a · bx in parentheses. The result is

(xy) = a · ay

ay· (bx)y = a

ay· (a · bx)y = a1−y · (f (x))y

7. (a) Since the ratio column is constant, the entries in theEarnings column are a geometric sequence. To gofrom 1 to 10 in two steps, the middle step must be10

12 ≈ 3.162

(b) k = 1012

1 = 1012 or

√10

(c) Multiply $1500 by√

10 to get $4743.42.8. (a) Since the ratio column is constant, the entries in the

Earnings column are a geometric sequence. You wantto go from 1 to 10 in 3 steps, so you multiply by 10

13

and j = 1013

1 = 1013 ≈ 2.154

(b) j = 1013

1 = 1013 or 3

√10

(c) Multiply $1500 by 1023 to get $6962.38

9. If x = 1, 1y = 1 for any real number y. If y = 0, x0 = 1,provided that x = 0. If x = −1, (−1)y = 1 if y is aneven integer or if y can be written in the form p

qwhere p

and q are relatively prime integers and p is even.

On Your Own

10. (a) 5x = 25 = 52, so x = 2(b) 5x = 5−11, so x = −11(c) 5x = 253 = (52)3 = 56, so x = 6(d) 5x = 0, so there is no solution.

11. Examples will vary.(a) Let x = 2 and y = 3, then f (x + y) = f (2 + 3) =

f (5) = 105 and f (x) · f (y) = f (2) · f (3) =102 · 103 = 102+3 = 105. Therefore, f (x + y) =f (x) · f (y).

(b) Let x = 4 and y = 3, then f (x − y) = f (4 − 3) =f (1) = 101 = 10 and f (x)

f (y)= f (4)

f (3)= 104

103 =104−3 = 101 = 10. Therefore, f (x − y) = f (x)

f (y).

(c) Let x = 3 and y = 2, then f (xy) = f (3 · 2) =f (6) = 106 and (f (x))y = (f (3))2 = (103)2 = 106

Therefore, f (xy) = (f (x))y .

12. (a) Solve 200 = 100 · 1.06x or 2 = 1.06x Estimate thevalue of x on your calculator. Since 1.0612 ≈ 2.0122,x must be a little less than 12. You can find x ≈ 11.90

(b) Solve 300 = 100 · 1.06x or 3 = 1.06x . Using yourcalculator, x ≈ 18.85 years.

(c) Solve 600 = 100 · 1.06x or 6 = 1.06x . Using yourcalculator, x ≈ 30.75 years.

13.

O�2�4 42x

abcdf

�

�

14. One possible answer is if N = 15. If f (x) = 2x = 15,then 2x = 15 = 3 · 5 = 21.5850 · 22.3219 =21.5850+2.3219 = 23.9069 then x = 3.9069.

Another possible answer is if N = 12 = 4 · 3. Iff (x) = 2x = 4 · 3 = 22 · 21.5850 = 22+1.5850 = 23.5850

then x = 3.5850.15. (a) f (2) = i2 = −1, f (3) = i3 = −i, f (5) = i5 = i

Since (−1) · (−i) = i, it is truethat f (2) · f (3) = f (5).

(b) The other rules also hold.

f (x − y) = ix−y = ix · i−y = ix

iy= f (x)

f (y)

f (xy) = ixy = (ix)y = (f (x))y

The graphs are translation images of each other. That isbecause, for each point (x, y) that is on the graph of a(x),there is a corresponding point with the same y-value onthe graph of each of the other functions. For example, thepoint (3, 27) is on the graph of a(x), but the point (1, 27)

is on the graph of c(x), since c(x) = 9 · 3x = 32 · 3x =3x+2. c(x) is the image of a(x) after a translation of twounits to the left.

.



16. (a) Plan1: 2 years because 25,000 + 3000 · 2 = 31,000Plan 2: 3 years because 25,000 · 1.083 = 31,492.80

(b) Plan 1: 5 years because 25,000 + 3000 · 5 = 40,000Plan 2: 7 years because 25,000 · 1.087 = 42,845.60

(c) Plan 1: 9 years because 25,000 + 3000 · 9 = 52,000Plan 2: 10 years because 25,000 · 1.0810 = 53,973.10

(d) Plan 1: 25 years because 25,000 + 3000 · 25 =100,000Plan 2: 19 years because 25,000 · 1.0819 = 107,893

17. f (x) = x3 is not an exponential function, it is apolynomial. In an exponential function, the variable is anexponent.

18. (a)

L(1) + L(2) = L(1 · 2)

L(1) + L(2) = L(2)

L(1) + L(2) − L(2) = L(2) − L(2)

L(1) = 0

L(2) + L(2) = L(2 · 2)

1 + 1 = L(4)

2 = L(4)

(b) L(x) = 6 = 2 + 2 + 2 = L(4) + L(4) + L(4) =L(4 ·4)+L(4) = L(16)+L(4) = L(16 ·4) = L(64),so, x = 64

(c) (i) n > 0

L(an) = L(a · a · a · · · a︸︷︷︸n copies

)

= L(a) + L(a) + · · · + L(a)︸︷︷︸n copies

= n · L(a)

(ii) n = 0

L(an) = L(a0)

= L(1)

= 0

n · L(a) = 0 · L(a)

= 0

So, L(an) = n · L(a)

(iii) n < 0Let n = −m where m > 0:

L(am · a−m) = L(am) + L(a−m)

L(a0) = m · L(a) + L(a−m)

L(1) = m · L(a) + L(a−m)

0 = m · L(a) + L(a−m)

−m · L(a) = L(a−m)

n · L(a) = L(an)

(d) For x > 0, we have L(x2) = 2 · L(x) = 1 = L(2).Since L is one-to-one, x2 = 2. Since x > 0, x = √

2.

19. Every year the sales is 1.15 times the previous year’s. Ifis the sales for the first year, then 5 years later (the sixthyear), the sales will be (1.15)5 · S. The desired ratio is

(1.15)5 = (1.15)8

(1.15)3≈ 3.059

1.521

which is barely over 2. The correct answer choice is B


20. (a)

(b) Since 4 = 22, think of this as 22x = 7. 22x = 22.807

2x = 2.807, x = 1.404(c) Since 8 = 23, 23x = 22.807, 3x = 2.807, x = .936(d) Since 16 = 24, 24x = 22.807, 4x = 2.807, x = .702(e) Since 1024 = 210, 210x = 22.807, 10x = 2.807,

x = 0.28121. (a) x = 1.984

(b) x = 0.504(c) x = 3.033(d) x = 0.330(e) x = 1.517(f) x = 0.659

. Exponential Functions One-to-One


1.

2. (a) L5(25) = 2 because 52 = 25(b) L7(1) = 0 because 70 = 1(c) L11(116) = 6 because 116 = 116

(d) L0.1(0.001) = 3 because 0.13 = 0.001(e) L3

(19

) = −2 because 3−2 = 132 = 1

93. (a) 1.1610

(b) 32

(c) 2.6610(d) 0.3390

4.

O

�2

�2

�4

�4

4

4

x

f�1f

y

8 05 ,

S

x = 2.807 You can find this on your calculator by“zooming” in. Since 23 = 8, start with values for x

that are a little less than 3.

We know that the domain of f (x) is the set of realnumbers and the range of f (x) is the set of positive realnumbers. Since L2(x) = f −1(x), the domain ofL2(x) = range of f(x) = {positive real numbers} and therange of L2(x) = domain of f(x) = {all real numbers}.

.

,



5. (a) The mean of f (3) and f (5) is 20 and f (4) = 16, sothe mean is greater.

(b) The mean of f (5) and f (7) is 80 and f (6) = 64, sothe mean is greater.

(c) The mean of f (−4) and f (−2) is 532 . f (−3) = 1

8 =4

32 , so the mean is greater.(d) The mean of f (x) and f (y) is greater than f

(x+y

2

),

because if you draw a line segment between any twopoints on the graph of 2x , the segment will lie abovethe curve. The mean is the midpoint of the segment,so it will be above (have a greater y-value) than thepoint on the curve with the same x-coordinate.

6. (a)

1 2 3 4�1�2�3�4

4

8

12

16

x

y

y � x2

y � 2x

O

(b) There are 3 solutions, since y = 2x and y = x2

intersect 3 times.7. No. Consider x = 200. 2002 = 40,000 while

1.06200 = 115,126. Starting at about x = 178, the valuesof 1.06x get larger than the values of x2.

8. (a) No. Starting at about x = 59, the values of 2x will belarger.

(b) No. Starting at about x = 1220, the values of 1.06x

will be larger.(c) Yes.

9. The value of b is approximately 1.4, and the coordinatesof the point are approximately (2.7, 2.7).

On Your Own

10. L2(x) is the inverse of f (x) = 2x and f (x) isone-to-one. So L2(x) must also be one-to-one.

11. L3(28) = x means that 3x = 28. 33 = 27 and 34 = 81,so L3(28) is between 3 and 4. The correct answer choiceis C.

12.

x L2(x)

1 02 13 1.58504 25 2.32196 2.58507 2.80748 39 3.170

13. (a) L8(8) = 1 because 81 = 8.(b) L8(2) = 1

3 because 813 = 3

√8 = 2.

(c) x L (x)

1 02 0.33333 0.5283

4 23

5 0.77406 0.86177 0.93588 19 1.0566

14. L2(x) = a means x = 2a

3 · L8(x) = b means L8(x) = b3 and

x = 8b3 = (23)

b3 = 2b

2a = 2b

a = b

L2(x) = 3 · L8(x)

15. To find the amount in her account after t years, use theexponential equation A(t) = $500 · 1.09t . When t = 8,the amount is $500 · 1.098 = $996.28. When t = 9, theamount is $500 · 1.099 = $1085.95. After 8 years, she isclosest to doubling her money.

16.Doubling Time

APR # Years to Double3% 234% 185% 146% 127% 108% 99% 8

10% 712% 6

17. (a) The Rule of 72 will gives the same number of yearsfor doubling the investment in all cases, except 3%,when it gives 24 years instead of 23 years.

(b) 72 ÷ 18 = 4, so the balance will double in 4 years.(c) Her money will double every 8 years, because

72 ÷ 9 = 8. In 40 years it will double 5 times. After 8years, she will have $500 · 2 = $1000. After 16 years,she will have $1000 · 2 = $2000. After 24 years, shewill have $2000 · 2 = $4000. After 32 years, she willhave $4000 · 2 = $8000. So, after 40 years she willhave $8000 · 2 = $16,000. Since her money has beenmultiplied by 2 five times, you can multiply500 · 25 = 16,000.


18. (a) 0.6826(b) 1.3652(c) 2.0478(d) 2.7304

8



19. (a) 2.8073(b) 1.4037(c) 0.9358(d) 0.7018(e) 0.2807

MATHEMATICAL REFLECTIONS

1. An exponential function is a function of the form

numbers as the domain and if a is positive, the realnumbers y > 0 as the range. (If a is negative, the range isthe real numbers y < 0.) If the base is 0 < b < 1 and a ispositive, then the function is decreasing for all x. This isalso the case if a is negative and b > 1. If b > 1 and a ispositive, the function is increasing for all x. This is alsothe case if a is negative and 0 < b < 1.

2. (a) The ratio between successive terms is 2, so to getfrom one term to the next, you multiply by 2. Tomake a complete recursive definition, though, youalso need to establish the first input-output pair.

g(x) ={

−5 if x = 0

2 · g(x − 1) if x > 0

(b) g(x) = −5 · 2x

(c) No, because the recursive function has only theintegers 0, 1, 2, . . . as its domain, and the domain ofthe closed-form function is the real numbers.

3. (a)

x h(x) ÷0 9 2

3

1 6 23

2 4 23

3 83

23

4 169

(b) h(23) = 9 · ( 23

)23 ≈ 0.0008.(c) h(x) could be an exponential function because the

entries in the table show a constant ratio betweensuccessive terms. It does not have to be anexponential function, though, because throughLagrange Interpolation you could find a polynomialfunction that matches this table completely.

4. (a) 12 = 3x2, so x2 = 4 and x ± 2.(b) x = 5 · 4− 1

2 = 5 · (22)−12 = 5 · 22·− 1

2 = 5 · 2−1 = 52

(c) 16 = 4 · 32x so 4 = 32x and 22 = (25)x or 22 = 25x .Because exponential functions are one-to-one, youknow that 2 = 5x and x = 5

2 .

(d) −5 = x · 2723 , so −5 = x · (33)

23 = x · 33· 2

3 = x · 9.Since −5 = 9x, x = − 5

9 .5. (a) If L3(1) = x, then 3x = 1 and x = 0.

(b) If L3(81) = x, then 3x = 81 = 34, so x = 4.(c) If L3(2) ≈ 0.6309, then 30.6309 = 2. To find L3(4),

you are looking for the solution to the equation3x = 4 and you can find this by squaring both sides of

the equation 30.6309 = 2. This gives you 32(0.6309) = 4,so L3(4) = 2(0.6309) = 1.2618. Similarly, cubeboth sides of the equation 30.6309 = 2 to get33(0.6309) = 8. This means that L3(8) = 3(0.6309) =1.8928.

(d) You know that L3(5) must be between 1 and 2,because 5 is between 31 and 32. Try different guessesfor L3(5), zeroing in on the correct value.

x 3x What next?1.5 5.1961 smaller1.4 4.6555 bigger1.45 4.9184 bigger1.46 4.9728 bigger1.47 5.0277 smaller1.465 5.0001 a little too big1.464 4.9947 too small1.4649 4.9996 1.4650 is better

6. Use the definition of the function to write an expressionfor the left side of the equation.

f (m) · f (n) = bm · bn

Then use the Laws of Exponents.

bm · bn = bm+n

Use the definition of the function.

bm+n = f (m + n)

This gives you the expression on the right side of theequation.

7. Exponential functions are strictly increasing or strictlydecreasing. This implies that every exponential functionis one-to-one, and all one-to-one functions have inverses.

8. You get 6% of the amount in the account as interest at theend of the year, so in effect, each year you have1.06 times as much as the year before. If you start with$1,000, you’ll have 1000 · (1.06)n after n years. After30 years, you have 1000 · (1.06)30 ≈ 5743.49 or$5743.49.

INVESTIGATION 8B

. Getting Started

For You to Explore

1. x log (x)

0 undefined1 02 0.30103 0.47714 0.60215 0.69906 0.77827 0.84518 0.90319 0.9542

10 1

TRANSFORMATIONS

8A

8 06

y = a · bx where a and b are real numbers, a �= 0, b > 0,and b �= 1. Examples will vary, but will have the real



2. Using the table, log (2) + log (3) = 0.3010 +0.4771 = 0.7781. Using your calculator and rounding tofour decimal places, log (2) + log (3) = 0.7782.

3. (a) 1.0792(b) 1.0792(c) 1.0792(d) x = 12

4. (a) You might estimate that log (15) =log (3) + log (5) = 0.4771 + 0.6990 = 1.1761

(b) log (24) = log (3) + log (8) = 0.4771 + 0.9031 =1.3802

(c) log (36) = log (6) + log (6) = 2 · log (6) =2 · 0.7782 = 1.5564. On the calculator, log (36) =1.5563

(d) log (63) = log (7) + log (9) = 0.8451 + 0.9542 =1.7993

5. log (MN) = log (M) + log (N)

6. (a) log (16) = log (2) + log (2) + log (2) + log (2) =4 · log (2) = 4 · 0.3010 = 1.2040. Using thecalculator, the result is 1.2041.

(b) log (32) = log (16)+ log (2) = 4 · log (2)+ log (2) =5 · log (2) = 5 · 0.3010 = 1.5050. Using thecalculator, the result is 1.5051

(c) log (64) = log (32)+ log (2) = 5 · log (2)+ log (2) =6 · log (2) = 6 · 0.3010 = 1.8060. Using thecalculator, the result is 1.8062.

(d) log (210) = 10 · log (2) = 10 · 0.3010 = 3.010.Using the calculator, the result is 3.0103.

(e) log (35) = 5 · log (3) = 5 · 0.4771 = 2.3855. Usingthe calculator, the result is 2.3856.

7. log (Mp) = p · log (M)

8. Since 18 = 2−3, log

(18

) = log (2−3) = −3 · log (2) =−3 · 0.3010 = −0.9030.

9. The domain is the set of positive real numbers. The rangeis the set of real numbers.

10. Take the log of both sides.

2x = 5

log (2x) = log (5)

x log (2) = log (5)

x = log (5)

log (2)

x = 0.69897

0.30103x = 2.3219

On Your Own

11. (a) 2(b) 3(c) 6(d) 10(e) −3

12. (a) 2(b) 3(c) 6(d) 10(e) This is undefined because −3 is not in the domain of

log (x).13. f (x) = log (x) and g(x) = 10x are inverse functions

because f (g(x)) = g(f (x)) = x. For example,f (g(2)) = f (102) = log (102) = 2 andg(f (2)) = g(log (2)) = 10log (2) = 2.

(d)log (x) + log (2) = log (5)

log (2 · x) = log (5)

2 · x = 5

x = 5

2

(e)

(f)log (x) = log (11) − log (4)

log (x) + log (4) = log (11)

log (4 · x) = log (11)

4 · x = 11

x = 11

4

15. log(

MN

) = log (M) − log (N)

16.

O

�2

�2

6

4

2

8

4 6 82

y

x

f

inverse of f

y

log (x) + log (7) = log (3)

log (7 · x) = log (3)

7 · x = 3

x = 3

7

14. (a) log(

54

)+ log (4) = log(

54 · 4

) = log (5) = log (x),so x = 5

(b) log(

73

)+ log (3) = log(

73 · 3

) = log (7) = log (x),so x = 7

(c) log(

617

)+ log (17) = log(

617 · 17

) = log (6) =log (x), so x = 6



17.

O

8

6

4

2

2 4 6 8x

f

inverse of fy

18. (a) log (AB) = log (A) + log (B) = 1.6 + 2.7 = 4.3(b) log (A2) = 2 · log (A) = 2 · 1.6 = 3.2(c) log

(1A

) = log (A−1) = −1 · log (A) = −1.6(d) log

(BA

) = log (B) − log (A) = 2.7 − 1.6 = 1.1(e) log (AB2) = log (A) + log (B2) = log (A)+

2 · log (B) = 1.6+ 2 · 2.7 = 1.6 + 5.4 = 7(f) log (

√A) = log (A

12 ) = 1

2 · log (A) = 12 · 1.6 = 0.8


19. • 0 and 1 • 1 and 2• 3 and 4 • 5 and 6

. Defining Logarithms


1. (a)

O

�2

�2

6

4

2

8

4 6 82

x

f

inverse of fy

(b) f −1(x) = 2x

2. (a) FALSE; 2(b) FALSE; 16(c) TRUE. Let the left side equal x. Then 5x = 10. Let

the right side equal y. Then 5y = 9. Since 10 > 9,5x > 5y and x > y and the statement is true.

(d) FALSE; 41 = 03. (a) 1

2

(b) 12 + 5

2 = 3(c) 3(d) 3

2(e) 3

2 + 12 = 2

(f) 2

4. logb (0) = x ⇔ bx = 0. bx = 0 only if b = 0, but b = 0,so logb (0) does not exist.

5. log2 (0.1) ≈ −3.322, but you really just need to concludethat it is less than zero for this exercise.log5

(15

) = −1, because 5−1 = 15 .

log73 (1) = 0, because 730 = 1.

log100 (10) = 12 , because 100

12 = 10.

log (99) has to be almost 2, because log (100) = 2.log4 (18) has to be more than 2, because log4 (16) = 2.log2 (18) has to be more than 4, because log2 16 = 4.This makes the correct order log2 (0.1), log5

(15

),

log73 (1), log100 (10), log (99), log4 (18), log2 (18).6. (a) b4 = 9

(b) b = 4√

9 = √3 or b ≈ 1.7321

7. Let x = 3log3 (75). Then use the definition of logb (M):

x = 3log3 (75)

log3 (x) = log3 (75) Definition of logarithm

x = 75

8.

9.4x = 25

log (4x) = log (25)

x log (4) = log (25)

x = log (25)

log (4)

x = 1.3979

.6021x = 2.322

10. You might start by looking at a series of logarithms suchas log (2000), log (200), log (20), and log (2). You willfind that

log (2) = 0.3010

log (20) = 1.3010

log (200) = 2.3010

log (2000) = 3.3010

All of these logarithms have the same mantissa, so themantissa tells you about the 2 that is part of each input.100.3010 ≈ 2. The ordinate (characteristic) in each case isthe power of 10 which is multiplied by the 2 to get theinput for the logarithm.

So if you know log (X) you can find X using the

8 07

3 �= 92 �= 4

�

�

(a) logb (2) = 1.35 ⇔ b1.35 = 2(b) logb (3) = 2.14 ⇔ b2.14 = 3(c) b3.49 = b1.35+2.14 = b1.35 · b2.14 = 2 · 3 = 6(d) b6.42 = b2.14·3 = (b2.14)3 = 33 = 27(e) logb (6) = x ⇔ bx = 6 But, we just showed that

b3.49 = 6. So, x = 3.49.(f) logb (27) = x ⇔ bx = 27 But, b6.42 = 27. So,

x = 6.42.

ordinate and mantissa: X = (10mantissa) · (10ordinate)

That’s fairly obvious, though, because10log (X) = 10ordinate+mantissa = X.



The ordinate, however, is a good indicator of the sizeof the number. If log (X) is positive and has ordinate O,then X has O + 1 digits to the left of the decimal. (If thelogarithm is negative, X < 1, and there are |O| zeroes tothe right of the decimal.) The mantissa, however, is reallyonly useful if you have to look up logarithms in a table.That means you can figure out the logarithm of anynumber X from a table that only lists the logarithms fornumbers from 0.0000 to 0.9999.

On Your Own

11. By definition logb (b) = 1 ⇔ b1 = b, which is true forany base b.

12. By definition, logb (1) = 0 ⇔ b0 = 1, which is true.13. (a)

by = M2

by = (bx)2

by = b2x

y = 2x

(b) M10 = (bx)10 = b10x or M10 = (M2)5 =(by)5 = b5y The missing exponent is 10x or 5y.

(c) By definition of logarithm, logb (M2) = y andlogb (M) = x. We found that y = 2x, sologb (M2) = 2 · logb (M)

14. Only (100, 2) is on the graph because f (100) =log (100) = 2. The correct answer choice is B.

15. (a) (b)

2 4 6 8 10

�2

4

6

8

10

(1, 0)(0, 1) (10, 1)

(1, 10)

xf

f –1

y

( , �1)(�1, )

101

101

(c) The domain of f is R+ and the range is R. Thedomain of f −1is R and the range is R+.

(d) f −1(x) = 10x

16. (a) log4 (18) = log (18)

log (4)= 2.085

(b) log2 (0.1) = log (0.1)

log (2)= −3.322

(c) log73 (1) = log (1)

log (73)= 0

(d) log1 (9) = log (9)

log (1)= log (9)

0 = undefined

(e) log3 (7) = log (7)

log (3)= 1.771

(f) log9 (49) = log (49)

log 9 = 1.771

17. (a) log2

(18

) = −3, because 2−3 = 18 .

(b) log2 (8) = 3, because 23 = 8.(c) log4 (16) = 2, because 42 = 16.(d) log4

(1

16

) = −2, because 4−2 = 116 .

(e) log1/2 (8) = −3, because(

12

)−3 = 8.

(f) log1/4

(1

16

) = 2, because(

14

)2 = 116 .

18. (a)

logb (M) = k

bk = M Definition of logarithm

(bk)−1 = (M)−1 Raise each side to the −1 power.

b−k = 1

M

logb

(1

M

)= −k Definition of logarithm

(b) Let logb M = k. Then bk = M . Taking reciprocals,1bk = 1

M, or

(1b

)k = 1M

. By the definition of

logarithm, log 1b

(1M

) = k = logb M .

19. The generalization would be logb (a) = logb2 (a2) Youcan prove this. First, let logb (a) = x. Then

logb (a) = x

bx = a

(bx)2 = a2

b2x = a2

(b2)x = a2

logb2 (a2) = x

logb2 (a2) = logb (a)

20. The base of a logarithm must be a positive number andnot 1. Therefore,

√3 is the only solution.


21. (a) 12

(b) 2(c) 3

2(d) 3

(f) 15

(g) 10.3010 ≈ 3.3

22. (a) 0.863(b) 1.863(c) 2.863(d) 6.863(e) 5.4771

2

(e) 5



. Laws of Logarithms


1. (a) logb (M2) = 2 · logb (M) = 2 · 2 = 4(b) logb (MN) = logb M + logb N = 2 + 5 = 7

(c) logbM3

N= logb M3 − logb N =

3 logb M − logb N = 3 · 2 − 5 = 6 − 5 = 1(d) logb (MN)3 = 3 logb MN =

3(logb M + logb N) = 3(2 + 5) = 3 · 7 = 21

(e) logb

√MN = logb (MN)

12 = 1

2 · logb MN =12 (logb M + logb N) = 1

2 (2 + 5) = 12 · 7 = 7

2

2. log2 3·log3 4·log5 6·log6 25 = log 3log 2 · log 4

log 3 · log 6log 5 · log 25

log 6 =log 4log 2 · log 25

log 5 = log2 4 · log5 25 = 2 · 2 = 4

3. log2 32log2 8 is not in the form logb

MN

, so you cannot apply the

rule logbMN

= logb M − logb N . In fact, log2 32log2 8 = 5

3 = 24. (a) Let bx = M and by = N so logb M = x and

logb N = y.Then,

bx

by= M

N

bx−y = M

N

logb

(M

N

)= x − y

logb

(M

N

)= logb M − logb N

(b) Think of MN

as (M) · ( 1N

).

logb

M

N= logb M − logb N

logb (M) ·(

1

N


logb M + logb

(1

N


logb M + logb N−1 = logb M − logb N

logb M − logb N = logb M − logb N

5. (a) Look at the Fundamental Law of Logarithms withN = 1.

logb M · 1 = logb M + logb 1

logb M = logb M + logb 1

This means that logb 1 must be a number that whenadded to logb M leaves it unchanged. So logb 1 = 0.

(b) Say that logb b = z. Convert this equation toexponential form. bz = b. By the Laws of Exponents,z = logb b = 1.

(c) Look at the Fundamental Law of Logarithms with

logb

(1

N

)· N = logb

1

N+ logb N

logb 1 = logb

1

N+ logb N

0 = logb

1

N+ logb N

− logb N = logb

1

N

6. (a) Since the cost grows by 3% = 0.03 each year, youwant to multiply by 1 + 0.03 = 1.03 each year. Thefunction is C(n) = 3.99(1.03)n

(b) C(n) = 20, so your equation is

20 = 3.99(1.03)n

log 20 = log (3.99(1.03)n)

log 20 = log 3.99 + log 1.03n

log 20 = log 3.

log 20 − log 3.99 = n · log 1.03

log 20 − log 3.99

log 1.03= n

n = 54.533 ≈ 55 years

7.

2 · 5x = 7

log (2 · 5x) = log 7

log 2 + log 5x = log 7

log 2 + x · log 5 = log 7

x · log 5 = log 7 − log 2

x = log 7 − log 2

log 5

x ≈ 0.778

8. Just follow the same steps as in the last problem.

a · bx = c

log (a · bx) = log c

log a + log bx = log c

log a + x · log b = log c

x log b = log c − log a

x = log c − log a

log b

OR

x = log ca

log b

9.

2 · 5x = 7 · 3x

log (2 · 5x) = log (7 · 3x)

log 2 + log 5x = log 7 + log 3x

log 2 + x log 5 = log 7 + x log 3

8 08

� 2.99 + n · log 1.03

x = log 3.5

log 5

M = 1N

.



x log 5 − x log 3 = log 7 − log 2

x(log 5 − log 3) = log 7 − log 2

x = log 7 − log 2

log 5 − log 3

x = log 72

log 53

x ≈ 2.452

10. Follow the same sequence of steps as in the last exercise.

a · bx = c · dx

log (a · bx) = log (c · dx)

log a + log bx = log c + log dx

log a + x log b = log c + x log d

x log b − x log d = log c − log a

x(log b − log d) = log c − log a

x = log c − log a

log b − log d

OR

x = log ca

log bd

11. Solve each equation for x.

6x = 35

log 6x = log 35

x log 6 = log 35

x = log 35

log 6

35x = 6

log 35x = log 6

x log 35 = log 6

x = log 6

log 35

Since log 35log 6 · log 6

log 35 = 1, the two results are reciprocals.12. (a) log x + log (x + 1) = log 6

log (x(x + 1)) = log 6

log (x2 + x) = log 6

x2 + x = 6

x2 + x − 6 = 0

(x + 3)(x − 2) = 0

x = −3 or x = 2

The solution x = −3 is not in the domain ofx (log −3 is not defined), so the only solution isx = 2.

(b) log (x2 + x) = log 6

x2 + x = 6

x2 + x − 6 = 0

(x + 3)(x − 2) = 0

x = −3 or x = 2

This time both solutions work because neither makesx2 + x negative or zero.

(c) log 2x − log x = log 2

log2x

x= log 2

log 2 = log 2

This equation will be true for all values in thedomain. The solution is x > 0.

(d) log2 (x − 3) + log2 (x + 3) = 4

log2 (x − 3)(x + 3) = 4

log2 (x − 3)(x + 3) = log2 16

(x − 3)(x + 3) = 16

x2 − 9 = 16

x2 = 25

x = ±5

x = −5, so the solution is x = 5.(e) log2 (x − 3) − log2 (x + 3) = 4

log2x − 3

x + 3= 4

log2x − 3

x + 3= log2 16

x − 3

x + 3= 16

x − 3 = 16x + 48

−15x = 51

x = −51

15= −17

5

This is not in the domain, x > 3, so there is nosolution.

13. You can set up an equation between two logarithms if youcan find two expressions that are equal to the samenumber. If log 2 = 0.3, then 5 log 2 = 5 · 0.3 = 1.5.

If log 3 = 0.5, then 3 log 3 = 3 · 0.5 = 1.5.Since both log expressions are equal to 1.5, you can set

them equal to each other. Use the rules of logarithms andalgebra to prove that 0 = 1.

5 log 2 = 3 log 3

log 25 = log 33

log 32 = log 27

32 = 27

32 − 32 = 27 − 32

0 = −5

0

−5= −5

−50 = 1

14. To find the number of years it takes an investment todouble, solve the equation 2A = A

(1 + r

100

)yfor y. (r

the interest rate, and A is the initial investment amount.)

But, �= −

is



2A = A(

1 + r

100

)y

2 =(

1 + r

100

)y

log 2 = log(

1 + r

100

)y

log 2 = y · log(

1 + r

100

)y = log 2

log(1 + r

100

)Now you can make a table comparing the value of 72

rto

log 2log(1+ r

100 )for interest rates from 6% to 12%.

r 72r

log 2log(1+ r

100 )

6 12 11.89577 10.2857 10.24488 9 9.00659 8 8.0432

10 7.2 7.272511 6. 5455 6.641912 6 6.1163

The results are pretty close!

On Your Own

15. log5 135 is the power of 5 that gives you 135. That is,5log5 135 = 135. Since 53 = 125 and 54 = 625,3 < log5 135 < 4. Similarly, log7 300 is the power of 7that gives you 300. Since 72 = 49 and 73 = 343,2 < log7 300 < 3. Therefore, log5 135 is larger.

16.

x log3 x

1 02 0.63093 14 1.26195 1.46506 1.63097 1.77128 1.89289 2

17. Since the log3 7 is the power of 3 that gives 7, log3 7 isless than 2 (32 = 9) but greater than 1(31 = 4).log9 81 = 2. Therefore, log3 7 + log9 81 is between 3and 4 (choice D).

18. You’re given that logb 297736 = 7 so you know thatb7 = 297736, or b = 7

√297736 ≈ 6.0531.

19. (a) log 6 = log (2 · 3) = log 2 + log 3 = a + b

(b) log 1.5 = log 32 = log 3 − log 2 = b − a

(c) log 27 = log 33 = 3 log 3 = 3b

(d) log 200 = log (2 · 100) = log 2 + log 100 = a + 2(e) log

√3 = log 3

12 = 1

2 log 3 = b2

20. From smallest to largest they arelog3 7 = 1.77124log8 40 = 1.77398log 60 = 1.77815

log12 83 = 1.77827log9 50 = 1.78044log2 3.44 = 1.78241

21. Since the bases are different, you cannot apply the rulelogb a + logb c = logb (ac). To find the correct value,evaluate each log separately. log 100 = 2 andlog2 8 = 3, so the correct value is 2 + 3 = 5.

22.

2x = 1

1,000,000

2x = 10−6

log 2x = log 10−6

x log 2 = −6

x = −6

log 2

x ≈ −19.932

23. (a)50 = 100 · 0.5h

1

2= 0.5h Divide both sides by 100.

log

(1

2

)= log 0.5h Take the log of both sides.

log 0.5 = h log 0.5

1 = h

(b)20 = 100 · 0.5h

1


log

(1

5

)= log (0.5h) Take the log of both sides.

log

(1

5

)= h log

1

2

log 5−1 = h log 2−1

− log 5 = −h log 2

log 5

log 2= h

2.322 ≈ h

(c)10 = 100 · 0.5h

1


log

(1

10


log 10−1 = h log 2−1

− log 10 = −h log 2

−1 = −h log 2

1

log 2= h

3.322 ≈ h

Notice that this answer is the sum of the previous twoanswers.



(d) c = 100 · 0.5h

c


log( c

100


log c − log 100 = h · log 0.5

h = log c − 2

log 0.5

(e) You can use the previous result:

h = log 1 − 2

log 0.5

or, since log 1 = 0,

h = −2

log 0.5≈ 6.6439

24. They are NOT the same. (log x)2 = log x · log x andlog x2 = log (x · x) = log x + log x.However, there are values for which (log x)2 = log x2.

(log x)2 = log x2

(log x)2 − 2 log x = 0

(log x)(log x − 2) = 0

So these two expressions will be equal if log x = 0,which means that x = 1, or when log x = 2, whichmeans that x = 100.

25. (a)

log2 x2 − 5 log2 x + 6 = 0

2 log2 x − 5 log2 x + 6 = 0

− 3 log2 x = −6

log2 x = 2

x = 22 = 4

(b)

(log2 x)2 − 5 log2 x + 6 = 0

(log2 x − 3)(log2 x − 2) = 0 Factor.

log2 x − 3 = 0 or log2 x − 2 = 0 Set each factor to 0.

log2 x = 3 or log2 x = 2 Solve for log2 x.

x = 23 = 8 or x = 22 = 4

(c)

log x = 1 + 6

log x

(log x)2 = log x + 6 Multiply by log x.

(log x)2 − log x − 6 = 0

(log x − 3)(log x + 2) = 0 Factor.

log x − 3 = 0 or log x + 2 = 0 Set each factor equal to 0.

log x = 3 or log x = −2

x = 103 = 1000 or x = 10−2 = 1

10026. Special case: let M = 1 in the original.

logb

(1

N

)= logb 1 − logb N = 0 − logb N = − logb N

27. It is true if M = N , because then both log M

log Nand M

Nare

equal to 1. Furthermore, their equality can be rewritten asN · log M = M · log N , which implies log (MN) =log (NM). Therefore, MN = NM . See the solution toLesson .1 , Exercise 13, for additional solutions withM = N .


28. Using the laws of logarithms,

log

(xy2

z

)= log (xy2) − log z

= log x + log (y2) − log z

= log x + 2 · log y − log z

The correct answer choice is B.29. Since the population is increasing at 2.5%, you want to

multiply by 1 + 0.025 = 1.025 each year. The function isP(n) = 10 · 1.025n where n is the number of years andP(n) is the population in millions after n years.

(a) 20 = 10 · 1.025n

2 = 1.025n

log 2 = log 1.025n

log 2 = n log 1.025

n = log 2

log 1.025

n ≈ 28.07

(b) 30 = 10 · 1.025n

3 = 1.025n

log 3 = log 1.025n

log 3 = n log 1.

n = log 3

log 1.

n ≈ 44.49

(c) 40 = 10 · 1.025n

4 = 1.025n

log 4 = log 1.025n

log 4 = n log 1.025

n = log 4

log 1.025

n ≈ 56.14

(d) 60 = 10 · 1.025n

6 = 1.025n

log 6 = log 1.025n

log 6 = n log 1.025

n = log 6

log 1.025

n ≈ 72.56

8 0

−

�=

025

.025

Mathematics III Solution Manual • Chapter , pages 2 98 6


(e) 120 = 10 · 1.025n

12 = 1.025n

log 12 = log 1.025n

log 12 = n log 1.025

n = log 12

log 1.025

n ≈ 100.63

(f) 240 = 10 · 1.025n

24 = 1.025n

log 24 = log 1.025n

log 24 = n log 1.025

n = log 24

log 1.025

n ≈ 128.70


Using algebra,

log2 (3x + 1) − log2 (2x − 3) = 2

log2(3x + 1)

(2x − 3)= log2 4

(3x + 1)

(2x − 3)= 4

(3x + 1)

(2x − 3)· (2x − 3) = 4 · (2x − 3)

3x + 1 = 8x − 12

−5x = −13

x = 13

5= 2.6

Using the graphing calculator,In y1, enter

log 3x+12x−3

log 2

In y2, enter

2

The point of intersection is (2.6, 2), so x = 2.6.(a) ln(0) is undefined. ln(1) = 0(b) One way to find b is to use your calculator. You can

determine that ln 2 ≈ 0.693147. Use thechange-of-base formula to get

ln 2 = logb 2 = log 2

log b≈ 0.693147

Try different values for b to eventually get thatb ≈ 2.71828 . . .

You can use a similar approximation technique,inputting different numbers into the LN functionuntil you find a number a so that LNa = 1.

3. (a)

O

�1

3

4

2

1

8 12 16 204

x

y

y � log2 x

y � √x

(b) There are 2 solutions. They are x = 16 and x = 4.4. No, after x ≈ 31622,

√x > log1.06 (x).

5. You can experiment with these functions as graphs or astables on your calculator to answer the questions:

(a) no(b) no(c) yes

6. You should graph each side of the equation separately,and look for the number of intersections.

(a) 2(b) 1(c) 2(d) 0(e) 2

7. Use your calculator and experiment with different valuesof b between 1.4 and 2, since when b = 1.4 there are twointersections, and when b = 2 there are no intersections.In fact, there is still no point of intersection whenb = 1.45. If b = 1.4447, the two points of intersectionare VERY close together. The actual value of b isapproximately 1.4447. The point of intersection is about(2.7183, 2.7183).

On Your Own

8. (a) Find 1.383 on the horizontal axis. Go up to the graphof f (x) = log2 x and across to the y-axis to read thesolution. log2 1.383 ≈ 0.468.

(b) Here you find 0.4 on the y-axis. Read across anddown to find the solution. log2 1.32 ≈ 0.4

9. The Laws of Logarithms say that logb ac = c logb a, soyou can say that log x2 = 2 log x. However, the graphsare not identical because the two functions have differentdomains. The domain of f (x) = log x2 is all realnumbers except zero. The domain of g(x) = 2 log x isonly positive real numbers. So, the graphs are not thesame.

8.09 Graphing Logarithmic Functions

1.

2.



10. (a)

O

2

4

8 12 164

n

D(n)

(b) D−1(n) = log1.06 n

(c)

O

4

8

12

16

2 4

n

D�1(n)

(d) Solve 1.06n = 10. Then n = log1.06 10, which isabout 40 years.

11. (a)

2

2 4 6 8

4

x

y

O

�4

�2

(b) Since y = logb x ⇔ by = x, as y gets bigger x getssmaller and closer to 0. For example, if b = 1

2 ,(12

)2 = 14 and

(12

)3 = 18 . This also means that as y

gets smaller, x gets bigger. For example,(

12

)−1 = 2

and(

12

)−2 = 4. If y gets smaller as x gets bigger, thefunction is decreasing.

12. Let y = g(x). Then

y = log8 x

8y = x Definition of logb x

(23)y = x

23y = x

3y = log2 x Definition of logb x

3g(x) = f (x) Substitute g(x) = y

and f (x) = log2 x.

Therefore, f (x) is three times as large as g(x).

13. (a)

2�2

�2

2

x

y

The graph is just the line with equation y = x,because y = log 10x = x log 10 = x · 1 = x.

(b)

O 2 4

2

4

x

y

y = 10log x

log y = log 10log x Take the log of both sides.

log y = log x · log 10 Law of Logarithms

log y = log x · 1

log x = log y

x = y

But, the domain is restricted because in the originaly = 10log x , the logarithm function cannot acceptnonpositive inputs. Therefore, the graph is only thepart of y = x where x > 0.

14. After graphing an example or two on your calculator, youmay be ready to believe that these graphs are horizontaltranslations of each other. In that case, there is somenumber t so that a point (x, y) on one of the graphs mapsto a point (x + t, y) on the other graph. That is what youwant to show in each of these cases.

(a) The graph of g(x) = 3x contains the point (m, 3m).There is a point on the graph of f (x) = a · 3x thathas the same y-coordinate (n, 3m). But since thispoint is on the graph of f (x), it must also satisfy theequation for the function, so 3m = a · 3n. Uselogarithms to solve this equation and show that thedifference m − n is a constant.

3m = a · 3n

m log 3 = log a + n log 3

m = log a

log 3+ n

m − n = log a

log 3



(b) The graph of g(x) = bx contains the point (m, 3m).There is a point on the graph of f (x) = a · bx thathas the same y-coordinate (n, bm). But since thispoint is on the graph of f (x), it must also satisfy theequation for the function, so bm = a · bn. Uselogarithms to solve this equation and show that thedifference m − n is a constant.

bm = a · bn

m log b = log a + n log b

m = log a

log b+ n

m − n = log a

log b

15. g(x) = f (x + 2) − 1, so the correct answer choiceis C.


16.

O 2�2

�2

�4

4

2

x

(a)

(e)

(d)(b)

(c)

y

The graphs of y = log3 (x − 3) and y = log3 (x) − 3 arenot the same. In the first, you subtract 3 from x and thentake the logarithm. This results in a horizontal translationof y = log3 x three units to the right. In the second, youtake the log of x and then subtract 3. This results in avertical translation of y = log3 x three units down.

.1 The Logarithmic Scale


1. The primes are less frequent among larger numbersbecause the values in the x

π(x)are getting larger. You are

told that out of all the numbers from 1 to x, about one inevery x

π(x)will be prime. If something occurs 1 in 22

times, that is less frequent than 1 in 8 times.

2. Here is a graph showing the first several points.

The graph appears to be a straight line.3. The graph appears to be a straight line, especially as the

values of x get larger.

With a logarithmic x-axis and a linear y-axis, thisindicates a logarithm function.

4. This scale is logarithmic because the values on thevertical axes that differ by the same amount are notrepresented by equal lengths. That is, the distancebetween 50 and 60 is greater than the distance between100 and 110.

5. (a) Consider any logarithmic function y = logb x withb > 0 and b = 1. Graphed with a logarithmic scaleon the x-axis, the base-10 logarithm outputs of thex-values are displayed. So you want to calculatelog x in terms of b and y, using the Laws ofLogarithms.

y = logb x

by = x

log by = log x

y log b = log x

y = 1

log b· log x

Since 1log b

is just some number, say B, the equationbecomes y = B log x, which is in linear form.

(b) The slope of that line is 1log b

.

8 0

�



6. (a) y = a · xb

log y = log (a · xb)

log y = log a + b log x

Let A = log a and b = B. The form of the equationis log y = A + B · log x, which is in linear form.

(b) The slope of the line is b.7. There are no negative numbers or zero on the logarithmic

axes. Therefore, the intersection point (0,0) does notappear. What the graph does show is that for any value ofx, 10x2 is 10 times x2, so the graphs are always the samedistance apart.

On Your Own

8. A function is one-to-one if it has an inverse, andf (f −1(x)) = f −1(f (x)) = x. In this case, you knowthat the inverse of f (x) = log2 x is f −1(x) = 2x . Youcan show thatf (f −1(x)) = log2 (2x) = x log2 2 = x · 1 = x andf −1(f (x)) = 2log2 x = x. Therefore, the function isone-to-one.

9. (a) (b)

The graph of y = x, shown in lighter points, lookslike a logarithmic graph would look on standard axes.This makes sense because log y = log x is what isbeing displayed. The graph of y = 1

xalso looks like a

logarithmic graph, but since it’s displaying log(

1x

),

as x gets larger, the y-values are decreasing.(c) Let v = log y. If y = x, then log y = log x or

v = log x. If y = 1x

, then log y = log(

1x

) =− log x or v = − log x. In the x-v plane, thesegraphs are reflections of each other over the x-axis(v = 0, which corresponds to y = 1).

10. (a) The bases are the same. log3 x = log3 7 ⇒ x = 7(b)

log3 x + log3 (x + 1) = log3 12

log3 x(x + 1) = log3 12

x(x + 1) = 12

x2 + x − 12 = 0

(x + 4)(x − 3) = 0

x = −4 or x = 3

But x = −4 because you cannot take the logarithmof a negative number. So, x = 3.

(c)

log (x2) = 4

x2 = 104

x2 = 10000

x = ±100

Both solutions work since x2 is positive.(d)

log5 (x + 1) − log5 x = 1

log5

(x + 1

x

)= log5 5

x + 1

x= 5

x + 1 = 5x

1 = 4x

1

4= x

11. (a) log4more than 2 because 42 = 16.

(b) log 27 is exactly 3.(c) log5 117 is the power of 5 that gives 117, so it is less

than 3 because 53 = 125.(d) log2 9 is the power of 2 that gives 9, so it is a little

more than 3 because 23 = 8.Therefore, choice(a) F(n) is exponential because you multiply by 2 to get

from one output to the next. The constant ratiobetween successive terms tells you that the functionmust be exponential.

�

17 is the power of 4 that gives 17, so it is a little

3

D is the largest.



(b) Exponential functions appear as straight lines whenyou plot them with a logarithmic scale on the y-axis.

13. Let y = kx. Then

xy = yx

xkx = (kx)x

(xk)x = (kx)x

xk = kx

xk−1 = k

Then to find y,

y = kx

y = k · k1

k−1

y = kk

k−1

Some ordered pairs that work:

Let k = 2. Then x = 21

2−1 = 2 andy = kx = 2 · 2 = 4. The ordered pair is (2, 4).

Let k = 3. Then x = 31

3−1 = 312 = √

3 and y = kx =3√

3. The ordered pair is (√

3, 3√

3).


14. (a)x log2 (x) log8 (x)

0 undef. undef.1 0 0

2 1 13

3 1.5850 0.5283

4 2 23

5 2.3219 0.77406 2.5850 0.86177 2.8074 0.93588 3 19 3.1699 1.0566

10 3.3219 1.1073

(b) If log8 (23) = 1.5079, then log2 (23) = 3(1.50794.5237.

15.

10�2

100

101

102

10�1

�10 �8 �6 �4 �2 0 2 4 6 8 10

y � 1.5x

y � 2x

y � 10 . 1.5x

y � 10 . 2x

16.

10�2

100

101

102

10�1

10010�110�2 101

y � x3

y � x2

y � 10 . x3

y � 10 . x2

MATHEMATICAL REFLECTIONS

1. (a)log

5

6+ log 6 = log

Apply the Fundamental Law of Logarithms.

log

(5

6· 6

)= log

Simplify.

log (5) = log x

Logarithmic functions are one-to-one.

x = 5

8B

) =

x

x

x = k1

k−1 (k �= 1)



(b)log3 2 + log3

1

2= log3 x


log3

(2 · 1

2

)= log3 x

Simplify.

log3 (1) = log3 x


x = 1

(c) log (10) · log (x) = log (30)

logb b = 1

1 · log (x) = log (30)

log (x) = log (30)


x = 30

(d)log2 (4) · log2

(1

2

)= log2 (x)

22 = 4 and 2−1 = 1

22 · (−1) = log2 (x)

−2 = log2 (x)

Convert the form.

2−2 = x

x = 1

4

2. log4 (2) < log4

(52

)< log4 (4), because 2 < 5

2 < 4, so12 < log4

(52

)< 1.

log2

(18

) = −3, because 2−3 = 18 .

log (0.1) < log (0.3) < log (1), because 0.1 < 0.3 < 1,so −1 < log (0.3) < 0.log7 (1) = 0, because 70 = 1.

log 12

(4) = −2, because(

12

)−2 = 4.So, the correct order is

log2

(1

8

), log 1

2(4), log (0.3), log7 (1), log4

(5

2

)3. (a) Find two consecutive powers of three, a and b, so

that a < 17 < b. a = 9 = 32 and b = 27 = 33 work,because 9 < 17 < 27. This means that j = 2 andk = 3, and 2 < N < 3.

(b) N = log3 17 = log 17log 3

(c) log3 (3 · 172) = log3 3 + 2 · log3 17 = 1 + 2N

4.

log (x − 20) + log (x − 50) = 3


log (x − 20 − ) = 3

103 = 1000, so log 1000 = 3

log (x − 20)(x − 50) = log 1000


1000

Multiply

x2 − 70x + 1000 = 1000

Simplify

x2 − 70x = 0

Factor

x(x − 70) = 0

Use the Zero Product Property.

x = 0 or 70

only solution.5. (a) If the graph of f (x) is increasing, you know that its

base, b, is greater than 1. If the graph is decreasing,0 < b < 1.

(b) f (x) = 0 says that logb x = 0. This impliesx = b0 = 1.

(c) f (x) must pass through (1, 0), because b0 = 1 andtherefore logb (1) = 0 for any base.

(d) The point (b, 1) must be on the graph because b1 = b

and therefore logb b = 1. Find an intersection

the base of the logarithm.6. Answers will vary, but should include the idea that you

can solve an equation of the form bx = c without usingtrial and error. Some students may also say thatlogarithms allow you to change a multiplication problemto an addition problem, or an expression involving anexponent to a product.

7.

.

)(x 50

(x − 20)(x − 50) =

Check each potential solution in the original equation.x = 0 results in negative inputs to a logarithmic function,so it is not a true solution to the equation. x = 70 is the

between the graph of f (x) and the line with equationy = 1. The x-coordinate of the intersection point is

A logarithmic scale is structured so that any two numbersa and b on the scale that are a distance d apart from eachother have the same ratio. (In a linear scale, any twonumbers a and b on the scale that are a distance d apartfrom each other have the same difference.) As you takeequal steps along a logarithmic scale, instead of addingyou multiply. Most logarithmic scales for graph paper arebuilt using base-10 logarithms. Only positive numberscan be shown on a logarithmic scale, and you might seevalues like . . . , 10−2, 10−1, 1, 10, 100, . . . labeling theaxes. It is convenient to use for graphing certain functionsinvolving logarithms or exponentials, “turning” thegraphs into ones that are linear, or close to it.

8. You want to solve the equation 10,000 = 1000(1 + 6

100

)for n.

10,000 = 1000

(1 + 6

100

)n

10 = (1.06)n

log 10 = n log 1.06

n = 1

log 1.06

39 5165



INVESTIGATION 8C EXPONENTIAL ANDLOGARITHMICFUNCTIONS

.11 Getting Started

For You to Explore

1. (a) At the end of the year, Danielle gets 100% interest onher $100 balance, for a total of $200.

(b) At 6 months, Danielle gets half of her 100% annualinterest on her $100 balance. Now her account has abalance of $150, and that’s the amount earning interestin the second 6 months of the year. When interest iscomputed again, she gets half of her 100% annualinterest on this $150 balance, for a total of $225.

(c) Here is a table showing the month, the interest addedto the account, and the new balance:

Month Interest Balance0 0 $1003 $100( 1

4 ) = $25 $1256 $125( 1

4 ) = $31.25 $156.259 $156.25( 1

4 ) = $39.06 $195.3112 $195.31( 1

4 ) = $48.83 $244.14

So at the end of the year, Danielle has $244.14.(d) Create a table like the one for part (c) for the case in

which Danielle gets 112 of her current balance as

interest at the end of each month. Her total balanceafter 12 months will be $261.30.

2. (a) If interest is compounded daily, Danielle gets$271.46. If it were compounded every minute oreven every second, she would get $271.83. It seemsas though there is a limit of about $271.83 toDanielle’s total yield. (All of these amounts arecomputed with the function in part (f).)

(b) a(n) = 100(1 + 1

n

)n, where 100 is the initial balance

in the account, n is the number of times the interest iscompounded, and 1

nis the 100% interest rate divided

by the number of times it is compounded.3. (a)

�4

�2

4

2

�4 42

(2, f(2))

(0, f(0))f(x) � 2x

�2 Ox

y

The slope of the secant is f (2)−f (0)

2−0 = 4−12 = 3

2 .

(b)

�4

�2

4

2

�4 42Ox

y

(0.5, f(0.5))(0, f(0))f(x) � 2x

The slope of the secant is f (0.5)−f (0)

0.5−0 =√2−10.5 = 2

√2 − 2 ≈ 0.828.

(c) Since the slope is positive and less than 0.828, a goodestimate might be 0.7.

4. (a)

�4

�2

4

g(x) � 4x (0, g(0))

(0.25, g(0.25))

�4 �2 42O

x

y

The slope of the secant is g(0.25)−g(0)

0.25−0 =1.4142−1

0.25 ≈ 1.66.(b) Since the slope is positive and a little less than 1.66, a

good estimate might be 1.5.5. One way to find a good approximation of the slope of the

tangent line at x = 0 is to take a point very close to x = 0and calculate the slope. For example, take(0.0001, f (0.0001)). The slope will be

b0.0001 − b0

0.0001 − 0= b0.0001 − 1

0.0001

Base b Slope of tangent to f (x) = bx at x = 02 0.6933 1.0994 1.3865 1.6098 2.07910 2.303

6. From the table, you can reason that b is between 2 and 3.If you use the slope formula b0.0001−b0

0.0001−0 = b0.0001−10.0001 and try

different values of b, a good estimate is b = 2.72.

It will take 40 years for the investment to be worth morethan $10,000.

8



On Your Own

7. (a)

�4

�2

4

2

�4 �2 4(1, f(1)) (2, f(2))

2

f(x) � log2x

O x

y

The slope of the secant is f (2)−f (1)

2−1 = 1−01 = 1.

(b)

�4

�2

4

2

�4 �2 42O x

y

(1, f(1)) (1.2, f(1.2))

f(x) � log2x

The slope of the secant is f (1.2)−f (1)

1.2−1 =0.263−0

0.2 ≈ 1.315.(c) Since the slope is positive and a little more than

1.315, a good estimate might be 1.4.8. (a)

�4

�2

4

2

�4 �2 2(1, g(1))

(2, g(2))

g(x) � log4x

4O x

y

The slope of the secant is g(2)−g(1)

2−1 = 0.5−01 = 0.5.

(b) Since the slope is positive and a little more than 0.5,a good estimate might be 0.7.

9. One way to find a good approximation of the slope of thetangent line at x = 1 is to take a point very close to x = 1and calculate the slope. For example, take(1.0001, f (1.0001)). The slope will be

log b1.0001 − log b1

1.0001 − 1= log b1.0001 − 0

0.0001

Base b Slope of tangent to f (x) = log bx at x = 12 1.4433 0.9104 0.7215 0.6218 0.48110 0.434

10. From the table, you can reason that b is between 2 and 3.If you use the slope formula log b1.0001−log b0

1.0001−1 =log b0.0001−0

0.0001 and try different values of b, a good estimateis b = 2.72.


11. (a)

(1 + x

2

)2 =(

2

0

)(1)2

(x

2

)0 +(

2

1

)(1)1

(x

2

)1 +(2

2

)(1)0

(x

2

)2

= 1 + x + x2

4

(b)

(1 + x

3

)3 =(

3

0

)(1)3

(x

3

)0 +(

3

1

)(1)2

(x

3

)1 +(3

2

)(1)1

(x

3

)2 +(

3

3

)(1)0

(x

3

)3

= 1 + x + x2

3+ x3

27

(c)

(1 + x

4

)4 =(

4

0

)(1)4

(x

4

)0 +(4

1

)(1)3

(x

4

)1 +(

4

2

)(1)2

(x

4

)2 +(4

3

)(1)1

(x

4

)3 +(

4

4

)(1)0

(x

4

)4


12. (a) f (1, 1) = 2, f (1, 2) = 2.25, f (1, 3) = 2.37037,f (1, 4) = 2.441406, f (1, 10) = 2.593742,f (1,100) = 2.704874, f (1,10000) = 2.718146

(b) f (2, 1) = 3, f (2, 2) = 4, f (2, 3) = 4.629630,f (2, 4) = 5.0625, f (2, 10) = 6.191736,f (2,100) = 7.244646, f (2,10000) = 7.387579

(c) f (0.05, 1) = 1.05, f (0.05, 4) = 1.050945,f (0.05, 12) = 1.051162, f (0.05,365) = 1.051267,f (0.05,10000) = 1.051271

= 1 + x + 3x2

8+ x3

16+ x4

256


.12 Compound Interest; the Number e


1. (a) • after 2 years:

100

(1 + 1

1

)2

= 100(4) = 400

• after 3 years:

100

(1 + 1

1

)3

= 800

• after t years:

100

(1 + 1

1

)t

= 100 · 2t

(b) • after 2 years:

100

(1 + 1

4

)8

= 596.05

• after 3 years:

100

(1 + 1

4

)12

= 1455.19

• after t years:

100

(1 + 1

4

)4t

= 100(1.25)4t

(c) At the end of 3 years, her balance would be100 · e3 ≈ $2008.55.

2. (a) • after one year:

$100

(1 + 0.06

1

)1

= $106.00

• after 2 years:

$100

(1 + 0.06

1

)2

= $112.36

• after t years:

$100

(1 + 0.06

1

)t

= 100 (1.06)t

(b) • after one year:

$100

(1 + 0.06

2

)2

= $106.09

• after 2 years:

$100

(1 + 0.06

2

)4

= $112.55

• after t years:

$100

(1 + 0.06

2

)2t

= $100 (1.03)2t = $100 (1.0609)t

(c) The $100 is the amount Jamie invested. The 1.005 is1 + 0.06

12 , 0.06 is the interest rate expressed as adecimal, and 12 is the number of times it iscompounded in a year (monthly). 36 = 12 · 3 is(number of times compounded in a year) (number ofyears).

3.

B = P

(1 + 0.05

n

)nt

4. (a) B = $100(1 + 0.05

4

)4·1 = $105.09

(b) B = $100(1 + 0.05

12

)12 = $105.12

(c) B = $100(1 + 0.05

365

)365 = $105.13

(d) B = $100(1 + 0.05

8760

)8760 = $105.135. Use your CAS to find these limits:

(a) B = limn→∞ $100(1 + 0.05

n

)5n ≈ $128.40

(b) B = limn→∞ $100(1 + 0.05

n

)10n ≈ $164.87

(c) B = limn→∞ $100(1 + 0.05

n

)20n ≈ $271.83

(d) B = limn→∞ $100(1 + 0.05

n

)40n ≈ $738.91

6. (a) $128.40 = $47.236 · e

(b) $164.87 = $60.652 · e

(c) $271.83 = $100 · e

(d) $738.91 = $271.83 · e

7. (a)

1

K= x

n

n · 1

K= n · x

nn

K= x

n = Kx

(b) (1 + x

n

)n =(

1 + 1

K

)Kx

(c) The two limits are the same because as n → ∞,K → ∞.

(d)

limK→∞

(1 + 1

K

)Kx

= limK→∞

((1 + 1

K

)K)x

=(

limK→∞

(1 + 1

K

)K)x

= ex

8. In the previous exercise, you saw that

limn→∞

(1 + x

n

)n = ex

So,

limn→∞ P

(1 + r

n

)nt = P(

limn→∞

(1 + r

n

)n)t

= P(er)t = Pert

8



On Your Own

9. One way to evaluate this to five decimal places is to carryout the calculations to a sufficient number of terms sothat there is no longer any change in the first five decimalplaces:

∞∑k=0

1

k! = 1

0! + 1

1! + 1

2! + 1

3! + 1

4! + 1

5! + 1

6! + 1

7!

+ 1

8! + 1

9! + · · · ≈ 2.71828

If you use your CAS, you get

∞∑k=0

1

k! = e

10. (a) Danielle is investing for one year at 100% APR.There is no compounding, so the amount in heraccount will be

1000

(1 + 1

1

)1·1= 2000

Originally Danielle invested $100 and her balancewas $200. Multiplying the investment by 10multiplies the balance at the end of the year by 10.

(b) The maximum amount will be when the interest iscompounded continuously:

B = $1000 · e1·1 = $1000 · e = $2718.28

11. (a) • B(3) = $1000 · e0.06·3 = $1000 · e0.18 = $1197.22• B(5) = $1000 · e0.06·5 = $1000 · e0.30 = $1349.86• B(t) = $1000 · e0.06·t

(b) Solve the equation:

2000 = 1000 · e0.06t

2 = e0.06t

log 2 = log e0.06t

log 2 = 0.06t log e

log 2

0.06 log e= t

11.55 ≈ t

It will take approximately 11.55 years for the balanceto double.

(c) Solve the equation:

4000 = 1000 · e0.06t

4 = e0.06t

log 4 = log e0.06t

log 4 = 0.06t log e

log 4

0.06 log e= t

23.1 ≈ t

It will take approximately 23.1 years for the balanceto become $4000.

12. Try an initial investment of $1000 for one year. Thebalance of the 6% account would be

B(1) = $1000

(1 + 0.06

1

)1

= $1000(1.06) = $1060

The balance for the account at 5.5% would be

B = $1000 · e0.055·1 = $1000 · e0.55 = $1056.54

So, the better investment is the account at 6% APR.13.

5 15 25

2

4

6

8f (t)

g(t)

35O

t

y

14. Take a point very close to (0, 1) and calculate the slope ofthe secant. For example, take(0.0001, e0.055·0.0001) = (0.0001, e0.000006). The slope is

e0.000006 − 1

0.0001 − 0≈ 0.055

15. g(t) = e0.055t = (e0.055)t ≈ 1.0565t

16. After one year, Jo has

$100

(1 + 0.04

4

)4

≈ $104.06

After one year, Eddie has

$100 · e.04 ≈ $104.08

The correct answer choice is A.


17. (a)

(1 + x

3

)3 =(

3

0

)(1)3

(x

3

)0 +(

3

1

)(1)2

(x

3

)1 +(3

2

)(1)1

(x

3

)2 +(

3

3

)(1)0

(x

3

)3

= 1 + x + x2

3+ x3

27

(b)

(1 + x

5

)5 ≈(

5

0

)(1)5

(x

5

)0 +(

5

1

)(1)4

(x

5

)1 +(5

2

)(1)3

(x

5

)2 +(

5

3

)(1)2

(x

5

)3

= 1 + x + 2x2

5+ 2x3

25

Mathematics III Solution Manual • Chapter , pages 2 98 7


(c)

(1 + x

10

)10 ≈(

10

0

)(1)10

( x

10

)0 +(

10

1

)(1)9

( x

10

)1 +(10

2

)(1)8

( x

10

)2 +(

10

3

)(1)7

( x

10

)3

= 1 + x + 9x2

20+ 3x3

25

(d)

(1 + x

n

)n ≈(

n

0

)(1)n

(x

n

)0 +(

n

1

)(1)n−1

(x

n

)1 +(n

2

)(1)n−2

(x

n

)2 +(

n

3

)(1)n−3

(x

n

)3

= 1 + x + (n − 1)x2

2!n + (n − 1)(n − 2)x3

3!n2

18. (a)

limn→∞

(n

1

)n

= limn→∞

n

n

= limn→∞ 1 = 1

(b)

limn→∞

(n

2

)n2

= limn→∞

n(n − 1)

2n2= lim

n→∞n2 − n

2n2

= limn→∞

1 − 1n

2= 1

2

(c)

limn→∞

(n

3

)n3

= limn→∞

n(n − 1)(n − 2)

3!n3

= limn→∞

n3 − 3n2 + 2n

3!n3= 1

3! = 1

6

(d)

limn→∞

(n

4

)n4

= limn→∞

n(n − 1)(n − 2)(n − 3)

4!n4

= limn→∞

n4 − 6n3 + 11n2 − 6n

4!n4= 1

4! = 1

24

(e)

limn→∞

(n

5

)n5

= limn→∞

n(n − 1)(n − 2)(n − 3)(n − 4)

5!n5

= limn→∞

n5 − 10n4 + 35n3 − 50n2 + 24n

5!n5

= 1

5! = 1

120

.13 Another Way to Find e


1. (a) 1 + 11! + 1

2! + 13! + · · · + 1

13! ≈ 2.7182818285. So, for14 terms, k = 13.

(b)(

1 + 1107

)107

≈ 2.71828169255, which gives e

correctly to six decimal places.2. (a) q(0.05) = 1.05125 and e0.05 = 1.05127. The percent

error is 1.05125−1.051271.05127 · 100 ≈ −0.002%.

(b) q(0.5) = 1.625 and e0.05 = 1.64872. The percenterror is 1.625−1.64872

1.64872 · 100 ≈ −1.44%.(c) q(−1) = 0.5 and e−1 = 0.36788. The percent error

is 0.5−0.367880.36788 · 100 ≈ 35.91%.

3. Solve ex = 2 on your CAS, or

ex = 2

log ex = log 2

xlog e = log 2

x = log 2

log e= log e2

x ≈ 0.693

4. (a) Since the slope at (0, 1) is one, a good start might bethe point where ex = 2, or (0.693, 2). Check the slopeof the tangent at that point by taking a nearby point:

e0.693 − e0.692

0.693 − 0.692≈ 1.9987 ≈ 2

(b) Solve ex = 3: x = 1.099. The point is (1.099, 3).Check the slope:

e1.099 − e1.098

1.099 − 1.098≈ 2.999 ≈ 3

5. (a)∑∞

k=0(−1)k

k! ≈ 0.36788(b) The summation gives the same rule as ex , where

x = −1. So it should equal e−1, and it does!6. (a)

�4 �2

4

2

6

8

42O

x

y

(b)

�8

�6

�4

�2

�4 �2 2 4Oxy8



(c)

6

8

4

2

�4 �2 42O

x

y

On Your Own

7. You can use the function defined by q(x) = 1 + x + x2

2 .q(0.03) = 1 + 0.03 + 0.0009

2 = 1.03045.8. (a) c(0.1) = 1.10516666667 and

f (0.1) = 1.10517091808. The percent error will be

c(0.1) − f (0.1)

f (0.1)· 100 = −0.00038%

(b) • c(0.2) = 1.22133333333 andf (0.2) = 1.22140275816. The percent errorwill be

c(0.2) − f (0.2)

f (0.2)· 100 = −0.00568%

• c(0.5) = 1.64583333333 andf (0.5) = 1.6487212707. The percent error will be

c(0.5) − f (0.5)

f (0.5)· 100 = −0.17516%

• c(1) = 2.666666666667 andf (1) = 2.71828182846. The percent error will be

c(1) − f (1)

f (1)· 100 = −1.89882%

• c(2) = 6.33333333 andf (2) = 7.38905609893. The percent error will be

c(2) − f (2)

f (2)· 100 = −14.28765%

(c) The graphs are very close together for small values ofx. As x gets larger, the graphs are farther apart.

(d) They are equal only when x = 0.9. Since

ex =∞∑

k=0

xk

k!1 + 2 + 4

2 + 86 + 16

24 + 32120 + · · · + 2k

k! + · · · =1+2+ 22

2! + 23

3! + 24

4! + 25

5! +· · ·+ 2k

k! +· · · = ∑∞k=0

2k

k! = e2

10. (a)ek = 3

log ek = log 3

k log e = log 3

k = log 3

log e= log e3

≈ 1.099

(b) m = log 5log e

= log e 5 ≈ 1.609

(c) n = log 15log e

= log e 15 ≈ 2.708

(d) p = log 45log e

= log e 45 ≈ 3.807(e) No solution. Since e > 0, er will never be negative.

11.

6

8

4

2

�4 �2 42Ox

y

Since e1.09861... = 3, this graph is just y = 3x .12. (a)

limn→∞

(1 − 1

n

)n

= limn→∞

(n − 1

n

)n

= limn→∞

(1n

n−1

)n

Let n = k + 1, n → ∞ ⇒ k + 1 → ∞ ⇒ k → ∞

= limk→∞

(1

k+1(k+1)−1

)k+1

= limk→∞

1(1 + 1

k

)k+1

= 1

e

(b) Show that it works for x = 0:

limn→∞

(1 + 0

n

)n

= e0 ⇒ 1 = 1

Now suppose x < 0.Let x

n= 1

−K⇒ n = −Kx where K > 0, and

n → ∞ ⇒ K → ∞. Then

limn→∞

(1 + x

n

)n = limK→∞

(1 + 1

−K

)−Kx

=(

limK→∞

(1 − 1

K

)K)−x

=(

1

e

)−x

= (e−1)−x

= ex

So,

limn→∞

(1 + x

n

)n = ex, x ≤ 0



13. Since ex = ∑∞n=0

xn

n! , it follows that e5 = ∑∞n=0

5n

n! . Thecorrect answer choice is B.


14. (a) 1 + 2

1= 3

(b) 1 + 2

1 + 1

6

= 19

7≈ 2.7143

(c) 1 + 2

1 + 1

6 + 1

10

= 193

71≈ 2.7183

(d) 1 + 2

1 + 1

6 + 1

10 + 1

14

= 2721

1001≈ 2.7183

. . . and they get progressively closer to e.

.14 The Natural Logarithm Function


1. Since 1024 = 210, ln 1024 = ln 210 =10 ln 2 = 10(0.6931) = 6.931

2. (a) ln 81 = ln 34 = 4 ln 3. So, p = 4 and M = 3.

(b) ln 3√

2 = ln 213 = 1

3 ln 2(c) ln 1

25 = ln 5−2 = −2 ln 5

3. (a) ln e2 = 2 ln e = 2(b) ln e10 = 10 ln e = 10(c) ln 1

e= ln e−1 = −1 ln e = −1

(d) eln 5 = 54. (a) ln 3

(b) m = ln 5(c)

ex ln 2 = eln 2x

= 2x

5. ax = eln ax = ex ln a = ekx , where k = ln a.6. (a)

2x = 7

ln 2x = ln 7

x ln 2 = ln 7

x = ln 7

ln 2≈ 2.807

(b) z = ln 123ln 5 ≈ 2.990

(c)

2 · 6x = 0.1

6x = 0.05

ln 6x = ln 0.05

x ln 6 = ln 0.05

x = ln 0.05

ln 6≈ −1.672

(d)

a · bx = c · dx

ln(a · bx) = ln(c · dx)

ln a + x ln b = ln c + x ln d

x ln b − x ln d = ln c − ln a

x(ln b − ln d) = ln c − ln a

x = ln c − ln a

ln b − ln d

= ln ca

ln bd

7. (a) (3, ln 3)

(b) (ln 3, 3)

(c) (2, e2)

(d) (e2, 2)

(e) ln g = 4 ⇒ g = e4

(f) ep = 4 ⇒ ln ep = ln 4 ⇒ p = ln 4

On Your Own

8. (a) ln 3 ≈ 1.099(b) ln 5 ≈ 1.609(c) ln 15 ≈ 2.708(d) ln 45 ≈ 3.807(e) ln 5

3 ≈ 0.511(f) ln(−5) is undefined.

9. (a) B(5) = $100 · e0.06(5) = $134.99B(10) = $100 · e0.06(10) = $182.21B(20) = $100 · e0.06(20) = $332.01

(b) Solve the equation:

100e0.06t = 200

e0.06t = 2

0.06t = ln 2

t = ln 2

0.06≈ 11.55

(c) Solve the equation:

100e0.06t = 400

e0.06t = 4

0.06t = ln 4

t = ln 4

0.06≈ 23.10

8



(d)

100e0.06t = B

e0.06t = B

100

0.06t = lnB

100

t = ln B100

0.06

t = ln 1000100

0.06 ≈ 38.3810. Use the result of Exercise 9. You want B to double from

100 to 200 and the interest rate to be p%:

t = ln 200100

0.01p= ln 2

0.01p

(a) ln 20.02 ≈ 34.7

(b) ln 20.03 ≈ 23.1

(c) ln 20.04 ≈ 17.3

(d) ln 20.05 ≈ 13.9

(e) ln 20.08 ≈ 8.7

(f) ln 20.01p

11. (a) 722 = 36 (actual 34.7), 72

3 = 24 (actual 23.1), 724 = 18

(actual 17.3), 725 = 14.4 (actual 13.9), 72

8 = 9 (actual8.7). It is a good approximation.

(b) The actual calculation involves dividing the ln 2 bythe interest rate expressed as a decimal.

12. This question asks for the value of t when

24,000 · e0.08t = 100,000

e0.08t = 100,000

24,000= 100

24

0.08t = ln

(100

24

)

t = 10.08 ln

(10024

) ≈ 17.84The decimal part is equivalent to about 10 months. The

correct answer choice is B.


13. (a) 8−54−3 = 3

(b) 4−38−5 = 1

3

(c) 2−110−2 = 1

8

(d) 10−22−1 = 8

(e) b−4a−0 = b−4

a

(f) a−0b−4 = a

b−414. One way to get these estimates is to define a function like

this one on your calculator:

s(x) = ln(x + 0.0000001) − ln x

0.0000001

Then you can just substitute the different values of x intothis function to get an estimate of the slope. You can also

use the limit capabilities of your calculator along with thelimit definition for the slope of the tangent.

(a) 1(b) 1

2

(c) 13

(d) 110

(e) 2(f) 3(g) The slope seems to be equal to 1

x.

.15 Analysis of f (x )=ex and g(x )=ln x


1. (a) To find the slope of the tangent line at some point(a, h(a)), take a nearby point, say(a + 0.0001, h(a + 0.0001)) and use your CAS tofind the slope of the secant line and the slope of thetangent line: 3ea+0.0001−3ea

0.0001 = 3ea So, the slope of thetangent at any point is equal to the y-value of thatpoint.

(b) Similarly,

e2(a+0.0001) − e2a

0.0001= 2e2a

So, the slope of the tangent at any point is equal tothe y-value of that point times 2.

(c) Similarly,

e5(a+0.0001) − e5a

0.0001= 5e5a

So, the slope of the tangent at any point is equal tothe y-value of that point times 5.

(d) Similarly,

e(ln 2)(a+0.0001) − e(ln 2)a

0.0001= ln 2e(ln 2)a

So, the slope of the tangent at any point is equal tothe y-value of that point times ln 2. Since ex ln 2 = 2x

you know that the slope of the tangent line at (a, 2a)

is ln 2 · 2a .2. (a) The slope is 20 · ln 2 = ln 2.

(b) The slope is 4 · ln 2 = ln 24 = ln 16(c) The slope is 51 · ln 5 = ln 55

7

5 (1, 5)

p(x) � 5x

3

1

�4 �2

�2

42O

x

y

8



3. (a) 50 ln 5 = ln 5(b) ln 8 = ln 23 = 3 ln 2(c) Base e, because e0 ln e = 1.(d) b = 1, but this is just a horizontal line:

y = 1x → y = 1.4. (a) This is the definition of logarithm.

(b) Take the ln of both sides and use the ruleln ab = b ln a:2y = x ⇒ ln 2y = ln x ⇒ y ln 2 = ln x

(c) Divide both sides by ln 2 to get y = ln xln 2 . Since

log 2x = y, log 2x = y = ln xln 2 .

5. (a) The slope is 3 times the reciprocal of the x-value.(b) The slope is the reciprocal of the x-value.(c) The slope is the reciprocal of the x-value times 1

5 .(d) The slope is the reciprocal of the x-value times 1

ln 2 .This function is log 2x.

6. (a) The slope at (1, 0) will be 11·ln 2 ≈ 1.443.

(b) The slope will be 14 ln 2 ≈ 0.361.

(c) The slope will be 15 ln 5 ≈ 0.124.

�1

�4

�2

75

(5, 1)

p(x) � log5x

31

4

2

O

y

x

7. (a) The slope will be 1ln 5 ≈ 0.621.

(b) The output for base 8 will be1

ln 8 = 1ln 23 = 1

3 ln 2 = 13

1ln 2 and 1

ln 2 is the output forbase 2.

(c) The slope of the tangent will be 1 if1

ln b= 1 ⇒ ln b = 1 ⇒ b = e.

(d) The slope of the tangent will be never be zero.

On Your Own

8. (a) Solve the equation:ex = 8 ⇒ ln ex = ln 8 ⇒ x = ln 8. So the point is(ln 8, eln 8) = (ln 8, 8).

(b) The equation of the tangent is

y − 8

x − ln 8= 8, or y = 8(x − ln 8) + 8

To find the y-intercept let x = 0:y = 8(0 − ln 8) + 8 = −8 ln 8 + 8 ≈ −8.636. They-intercept is negative.

9. You are looking for a point (a, ea) where the tangent lineto the graph of f (x) = ex has a y-intercept of 0. Firstwrite the equation of the tangent line:

y − ea

x − a= ea, or y = ea(x − a) + ea = eax − aea + ea

If the y-intercept is 0, then−aea + ea = 0 ⇒ ea(−a + 1) = 0 ⇒ ea �= 0, a = 1.

So, the point is (1, e) and the slope is e. The equation ofthe tangent is

y − e

x − 1= e, or y = e(x − 1) + e = ex

10. The rule is slope = b · p(x) = Abebx

11. In Exercise 5, you saw that if q(x) = A ln Bx, the rulethat can give the slope of the tangent line is slope = A

x.

12. In Exercise 2, you saw that if r(x) = bx , the rule that cangive the slope of the tangent line is slope = bx · ln b.

13. In Exercise 6, you saw that if s(x) = log bx = ln xln b

, therule that can give the slope of the tangent line isslope = 1

x ln b.

14. In Exercise 2, you saw that if m(x) = bx , the slope of thetangent is bx · ln b. In Exercise 6, you saw that ifm(x) = log bx then the slope of the tangent is 1

x ln b. In

both cases, the slope involves ln b. In the first case, sinceit is related to ex , you multiply by the y-value of thepoint, or bx . In the second case, you can relate to ln x, soyou take the reciprocal of x ln b.

15. Starting with the rule p(x) = Aebx , the slope at (d, c) isbc. The inverse function is p−1(x) = 1

bln( x

A), and the

slope at (c, d) is 1bc

. They are reciprocals.16. The slope is 5c · ln 5, so the equation could be written as

y − 5c

x − c= 5c · ln 5

y − 5c = 5c · ln 5(x − c)

y = 5c · ln 5(x − c) + 5c

The correct answer choice is D.

8C MATHEMATICAL REFLECTIONS

1. To find the amount in Adam’s account after 5 years withquarterly compounding, calculate

500

(1 + 0.06

4

)5·4= 673.43.

For continuous compounding, calculate

limn→∞ 500

(1 + 0.06

n

)5·n= 674.93

So the difference is about $1.50.2. (a) The factorial definition of e is

e =∞∑

k=0

1

k! .

The first five terms in the sum are1

0! + 1

1! + 1

2! + 1

3! + 1

4! =1

1+ 1

1+ 1

2+ 1

6+ 1

24=

24

24+ 24

24+ 12

24+ 4

24+ 1

24=

65

24≈2.708333



(b) e−2.708333e

· 100 ≈ 0.365997(c) You can define the function d(n) = (

1 + 1n

)non your

calculator and look at its value for different inputs.d(50) ≈ 2.691588, which isn’t close enough.d(100) ≈ 2.704814, which is closer.d(150) ≈ 2.709276, which is a better estimate thanthe one in part (a). The first value of d(n) that gives abetter estimate than part (a) is d(136) ≈ 2.708355.

3.

8000 = 5000 · e0.07t

1.6 = e0.07t

ln 1.6 = 0.07t

t = ln 1.6

0.07t ≈ 6.7

That’s about 6 years, 9 months.4. The slope of the tangent to the graph of y = 3 ln x at a

point (a, 3 ln a) is 3a

, so the slope of this tangent is 32 . An

equation for the line is 32 = 3 ln 2−y

2−x, or

y = 32x − 3(1 − ln 2).

5. g(x) = ex·ln 5. The slope of the tangent to the graph of g

at (2, 25) is 52 ln 5 = 25 ln 5, so an equation for thetangent to the graph of g is 25 ln 5 = 25−y

2−xor

y = (25 ln 5)x + 25 − 50 ln 5.6. The more frequently you compound the interest, the

greater your yield, but there is an upper bound. As thenumber of times you compound the interest increases, thelimit is continuous compounding, so that the total amountb that an investment of P dollars at interest rate r over aperiod of t years is worth is b = Pert .

7. Answers will vary. Key reasons include noticing theupper bound for interest compounded more and morefrequently, or perhaps looking for a function in which theslope of the tangent to the graph of the function at anypoint is equal to the y-value of the function.

8. Any exponential function h(x) = bx , for b > 0, can berewritten as h(x) = ex ln b. Any logarithm functionj (x) = log b(x), for b > 0, can be rewritten asj (x) = ln x

ln b.

CHAPTER REVIEW

1. (a) You are looking for an equation in the formf (x) = a · bx . Write f (2) = a · b2 = 9

2 andf (3) = a · b3 = 27

2 . Solve each equation for a:

a =92

b2

a =272

b3So,

92

b2=

272

b3

9

2· b3 = 27

2· b2

9b3 = 27b2

9b3 − 27b2 = 0

9b2(b − 3) = 0

So b = 0 or b = 3. But b > 0, so b = 3. Substitute tofind a:

f (2) = a · 32 = 9

2−→ 9a = 9

2−→ a = 1

2

Therefore f (x) = 12 (3)x .

(b)

O 2�2

6

4

2

8

10

12

14

x

y

4

(0, 1 )

(2, 9 )

(3, 27)

2

2

2

(a)

x g(x) ÷0 −2 −20

−2 = 10

1 −20 −200−20 = 10

2 −200 −2000−200 = 10

3 −2000 10

To find the next term, you multiply by 10. So,

g(x) ={

−2 if x = 0

10 · g(x − 1) if x > 0

(b) A closed-form definition will be in the formg(x) = a · bx . g(0) = −2 = a · b0. So, a = −2.Substitute to find b: g(1) = −20 = −2 · b1 −→ b =10. The closed-form definition is

g(x) = −2(10)x

(a) If L4(1) = x, then 4x = 1 and x = 0.(b) If L4(64) = x, then 4x = 64 = 43, so x = 3.(c) If L4

(116

) = x, then 4x = 116 = 1

42 = 4−2, sox = −2.

.

3.

4 You can find the exact or approximate value of eachnumber by using your calculator and the change-of-base

.2

(c) g(4) = −2(10)4 = −2(10,000) = −20,000 andg(6) = −2(10)6 = −2(1,000,000) = −2,000,000.



formula or by using the definition of logb (x).

log2 (16) = 4

log 5 ≈ 0.699

log3 (8) = log 8

log 3≈ 1.893

log4

(1

8

)= −3

2

log 12

(16) = −4

Arrange the numbers from smallest to largest:

log 12

(16), log4

(1

8

), log 5, log3 (8), log2 (16)

. (a) log x − log 5 = log 15

log(x

5

)= log 15

x

5= 15

x = 15 · 5 = 75

(b) log2 (x − 1) + log2 (x + 1) = 3

log2 ((x − 1)(x + 1)) = 3

log2 (x2 − 1) = 3

x2 − 1 = 23

x2 − 1 = 8

x2 = 9

x = ±3

−3 cannot be a solution because you cannot take thelog of a negative number and x − 1 < 0 if x = −3.The only solution is x = 3.

(c) 5x = 7

log (5x) = log 7

x log 5 = log 7

x = log 7

log 5

x ≈ 1.209

(d) 3 · 2x = 20

2x = 20

3

log 2x = log

(20

3

)

x log 2 = log

(20

3

)

x = log(

203

)log 2

= log 20 − log 3

log 2

x ≈ 2.737

5

. (a)

2

4

x

f

g

y

O

�2

(b) The graphs intersect in one point. There is onesolution.

(c) Find the point of intersection: x ≈ 1.36

7. (a) e0.08 ≈ 1 + 0.08 + 0.082 + 0.08

6 + 1.08329(b) e0.08 ≈ (1 + 0.08

10 )10 ≈ 1.08294(c) e0.08 ≈ 1.08329

8. (a) Let t = 5:

B = 1000e0.055t

= 1000e0.055.5

= 1000e0.275

= 1316.53

Let t = 10 :

B = 1000e0.055t

= 1000e0.055.10

= 1000e0.55

= 1733.25

(b) LetB = 2000You want to solve the equation 2000 = 1000 ·

e0.055t for t . Divide both sides by 1000. Solve theequation for t by taking the ln of both sides:

2000 = 1000 · e0.055t

2 = e0.055t

ln 2 = ln(e0.055t )

ln 2 = 0.055t

ln 2

0.055= t

12.603 ≈ t

Repeat usingB = 10,000 :

10,000 = 1000e0.055t

10 = e0.055t

ln 10 = ln (e0.055t )

ln 10 = 0.055t · ln e

ln 10 = 0.055t

ln 10

0.055= t

41.865 ≈ t

6

2 3



9. (a) (2, f (2)) = (2, e2). The slope of the tangent line ise2, and the equation of the tangent line is

y − e2

x − 2= e2 ⇒ y − e2 = e2(x − 2) ⇒ y = e2x − e2

Since e2 ≈ 7.38906, you can write the equation asy = 7.38906x − 7.38906.

(b) (2, g(2)) = (2, ln 2). The slope of the tangent line is12 , and the equation of the tangent line is

y − ln 2

x − 2= 1

2= y−ln 2 = 1

2(x−2) = y = 1

2x−1+ln 2

Since ln 2 ≈ 0.693147, you can write the equation asy = 0.5x − 0.306853.

(c) Since the slope at the point (a, ea) is ea , you wantea = 3:

ea = 3

ln ea = ln 3

a = ln 3 ≈ 1.09861

The point is (ln 3, 3) ≈ (1.09861, 3).(d) Since the slope at the point (a, ln a) is 1

a,you want

1a

= 13 , So, a = 3. The point is

(3, ln 3) ≈ (3, 1.09861).

CHAPTER TEST

Use the Laws of Logarithms. The only choice that doesnot work is C because log (a · b) = log (a) + log (b) notlog (a) · log (b).By definition, y = log3 (x) means 3y = x. The onlychoice that solves this equation is (9,2) because 32 = 9.The correct choice is C.

. (a)

x h(x) ÷0 6 3

6 = 12

1 3323 = 1

2

2 32

3432

= 12

3 34

3834

= 12

4 38

(b)

h(x) ={

6 if x = 012 · h(x − 1) if x > 0

(c) Write a function in the form h(x) = a · bx . Sinceh(0) = 6, a · b0 = 6 and a = 6. Since h(1) = 3 andh(1) = 6 · b1, b = 3

6 = 12 . The closed-form

definition is

h(x) = 6

(1

2

)x

(d) h(8) = 6(

12

)8 = 6 · 1256 = 3

128

$2250

(1 + 0.07

4

)6·4= $2250 (1.0175)24

≈ $3412.00

The correct answer is A.

1.

2.

3.

4

(a) ex = 5 ⇒ ln ex = ln 5 ⇒ x = ln 5 ⇒ x ≈ 1.609(b) ln x = 2 ⇒ eln x = e2 ⇒ x = e2 ⇒ x ≈ 7.389Theorem . says that the point where the graph off (x) = ln x has a tangent line with slope 5 is

(15 , ln 1

5

).

5.

6.

7. (a)(i)

x f (x)

−1 14

0 112 2

1 42 16

(ii)

x g(x)14 −1

1 0

2 12

4 1

(b)

O 2�2

�2

4

2

4

xg

y

f

(c) The graph is increasing from left to right.(d) The graph is increasing from left to right.

8. (a) By definition, N = log 28 N = 28. Since16 < 28 < 32 −→ 24 < 2N < 25, N must bebetween j = 4 and k = 5.

(b) log (7) = 2

(284

) = log2 (28) − log2 (4) = N − 2

(c) log2 (28) = log (28)

log (2)≈ 4.807

.

M = 10,000(1 055)n

9

2 ( ) if 2

log2

Write an exponential equation:

.

8 7



where M is the amount of money in the account after n

years. You want to know when M = 15,000. Uselogarithms to solve the equation:

10,000(1.055)n = 15,000

1.055n = 1.5

log (1.055n) = log (1.5)

n log (1.055) = log (1.5)

n = log (1.5)

log (1.055)

≈ 7.573

So, it will take 8 years for the money to reach $15, 000because the interest is applied annually.

1 .

Amount in15 years

Time toDouble

InitialInvestment APR

$95,000.19

$48,217.827.7

years9%

8%

17.33%

11.55%

$107,903.80

$127,841.57

$12,500

$32,500

$9,500

$16,800

8.66years

4years

6years

0



Documents

000200010271835410 CH08 p000-000 · There are many possible solutions. For example, ... CME_Int_L3_CSM_CH08 03/03/2014 14:43 Page 254 (a perfect square) or 8 because 72 ÷8 = 9 (a