4.4 COMPUTATIONS WITH LOGARITHMIC AND …

[Pages:10]4.4 Computations with Logarithmic and Exponential Functions

227

Exercises 56?57 Maximum Value (a) For what value(s) of x is y equal to 36 (one decimal place)? (b) What value of x will give a maximum value of y? What is the maximum value? (Hint: The window 0, 300 0, 60 should give you a start.)

56. y 4 xe0.01x

57. y 6 x ? 30.01x

58. Maximum Concentration The concentration C of a

drug in the bloodstream at t minutes after injection is

given by

C 0.036te0.015t mgcm3.

(a) In how many minutes will the concentration reach 0.6 mg/cm3?

(b) How many minutes after injection will the concentration be the greatest? What is the maximum concentration? See Example 11.

59. True or False Draw graphs to support your answer.

Assume that L is a line. (a) If L and the graph of y ln x intersect at two

points, then the slope of L must be positive. (b) If L and the graph of y ex intersect at two

points, then the slope of L must be positive.

60. Explore For what integer values (positive and nega-

tive)

of

c

will

the

graphs

of

y

1

x c

and

y

ln

x

intersect at (a) exactly one point? (b) two points?

61. Explore For what integer values (positive and negative) of c will the graphs of y cx 3 and y 2x 6 intersect at (a) exactly one point? (b) two points?

62. Explore What is the smallest prime number c for which the graph of y cx 5 will intersect the graph of y 3x 5 at exactly two points?

63. Your Choice Give a formula for a linear function f (with nonzero slope) that satisfies the specified conditions. (a) The graphs of f and y ln x intersect in Quadrant I and Quadrant IV. (b) The graphs of f and y ex intersect in Quadrant I and Quadrant II.

64. Your Choice From the family of functions f x c lnkx, where c and k are nonzero constants, select c and k so that f satisfies the specified condition. (a) The graph of f intersects the graph of y 2x 4 at two points. (b) The graphs of f and of y x 4 intersect in Q II.

Exercises 65?66 Is It a Function? observe when you graph the equation. 65. y lnx 2 2x 3 66. y lnx 3 ln2 x

Explain what you

4.4 C O M P U T A T I O N S W I T H L O G A R I T H M I C AND EXPONENTIAL FUNCTIONS

Galileo's observation that all bodies accelerate equally in the Earth's gravity is counterintuitive precisely because it is usually wrong. Everybody knows that a lump of coal falls faster than a feather. Galileo's genius was in spotting that the differences which occur in reality are an incidental complication caused by air resistance, and are irrelevant to the properties of gravity as such.

P. W. C. Davies

Because logarithms are exponents, evaluation in exact form is possible only in special cases. We can, for example, evaluate log3(93) in exact form because 93 is a power of the base 3:

93

352,

so

log3

93

5 2

.

More generally, we need assistance to approximate logarithms. This section covers the use of calculators to evaluate logarithmic and exponential functions to any base. All scientific calculators are programmed to evaluate the natural

228

Chapter 4 Exponential and Logarithmic Functions

I was in the ninth grade

of Powell Junior High School in Washington, D.C. I was doing very poorly in my first course in algebra. To be precise, I was flunking. Later on, after recovering from my poor start in algebra, I began to get top marks. I was good in math and science in high school.

George B. Dantzig

exponential function, f x ex, and its inverse, the natural logarithm function f x ln x. As we will see, these functions are sufficient to handle calculator

evaluation for exponential and logarithmic functions with any base.

Change of Base and Evaluating Logarithms in Other Bases

As observed in the previous section graphing calculators have an exponentiation

key, or yx , that allows us to evaluate exponential expressions or to graph 57 57

exponential functions for any given base. In contrast, there is no built-in logarithm

key that directly evaluates logarithms for any bases except e LN and 10 log .

57

57

Fortunately, there is a simple change-of-base formula that allows us to evaluate any

logarithmic function by means of the natural logarithm function, ln x. To evaluate

log3 4, we can express the relationship y log3 4 in exponential form, (by equivalence Equation (1)), then apply the natural logarithm function to both sides and

solve for y. The same steps work for any base b, as follows:

y log3 4 3y 4 ln 3y ln 4 y ln 3 ln 4 y ln 4

ln 3

y logb c by c ln by ln c y ln b ln c y ln c

ln b

By EQ1 Applying ln function By L3

Solving for y

Thus

we

have

log3

4

ln ln

4 3

1.2619,

and

we

have

a

general

formula

for

evaluat-

ing any logarithmic function.

Change-of-base formula

For

any

positive

real

numbers

c

and

b

where

b

is

not

1,

logb

c

ln ln

c .

b

The change-of-base formula allows us to evaluate logarithmic functions for any base, including base 10, so that 5l7 og is not really necessary.

EXAMPLE 1 Evaluating logarithms Find an approximation rounded off to four decimal places.

(a) log5 0.43 (b) log81 3

Solution Use the change-of-base formula.

(c) log 79.442

(a)

log5

0.43

ln 0.43 ln 5

0.5244.

(b)

log81

3

ln1 3 ln 8

0.4833.

(c) With no base shown, log 79.442 refers to the common logarithm (base 10). Use

log directly if your calculator has such a key, or use the change-of-base 57 formula.

log 79.442 ln 79.442 1.9001. ln 10

Check each of the above computations using your calculator.

4.4 Computations with Logarithmic and Exponential Functions

229

There are many occasions when we have functions given by two different formulas and we want to determine whether the functions are identical. A graphing calculator can be very helpful in this regard, and there are at least three convenient methods.

TECHNOLOGY TIP Graphing identical functions

We want to determine whether two functions, f and g, are identical.

? Method 1 Plot the graphs of y f x and y gx on the same screen. The advantage and disadvantage of this method is that you see only one graph. Differences in domain may not be apparent. To check, trace along the curve, using the up or down arrows to jump from one curve to the other, and watch the y-coordinates.

? Method 2 Translate one graph up or down by some constant, say 1 or 0.5. That is, plot the graphs of y f x and y gx .5 on the same screen. If the functions are identical, the graphs will differ by the same amount all the way across the screen.

? Method 3 Shift the graph of f g so that the difference is visible on the screen. Plotting y f x gx 1 will yield the horizontal line y 1, which can also be checked by tracing (or replace 1 by any other constant).

EXAMPLE 2 Verifying the change of base formula Use graphs to support the claim that the functions f x log x and gx ln xln 10 are identical.

Solution Following the suggestions in the Technology Tip above, we enter Y1 LOG X and Y2 LN X/LN 10. We can graph both Y1 and Y2 on the same screen and see a single logarithm function, or we can graph Y1 and Y2 1 (for Method 2), or Y3 Y1 Y2 1 (for Method 3). By whichever method we choose, the calculator shows that, at least to calculator accuracy, the functions are identical.

Using Inverse Function Identities

Restating the inverse function identities in terms of the natural exponential function and the natural logarithmic function is useful as a reminder of relations that can simplify much of our work.

If f is the natural exponential function, f x ex, then f 1x ln x. Since f f 1x x for all x in the domain of f 1 and f 1 f x x for all x in the domain of f , we have two identities.

Inverse function identities

eln x x for all positive numbers x.

(1)

ln ex x for all real numbers x.

(2)

230

Chapter 4 Exponential and Logarithmic Functions

HISTORICAL NOTE I N V E N T I O N O F L O G A R I T H M S

As the need for more accuracy in

for accuracy N 10,000,000

trigonometric computations grew

and calculated a hundred terms in

(see the Historical Note,

a geometric sequence, successively

Trigonometric Tables in Section 5.3), so did the need for

subtracting

1 10,000,000

of

each

number

from the one before, and rounding

better ways to do the arithmetic.

each to 14 digits.

Logarithms have been called "the

This produced one table of

most universally useful

exponents. If he had simply

mathematical discovery of the

continued with this sequence, it

seventeenth century." They

would have required years of

significantly reduced the time

calculation just to get from

required to perform computations

10 million to 5 million, producing

and may have been as important for

an unusable table with nearly 7

the exploration of the globe as any

million entries. Napier's genius lay

improvement in marine technology

in his construction of other tables

in two hundred years.

to allow interpolation between

One basic idea motivated the

numbers. Rather than millions of

development of logarithms: to

entries, his second table had only

multiply powers of the same base,

50 entries, and the third had fewer

simply add exponents. For example,

than 1500. A user would locate a

to multiply 16 by 64, use tables to

pair of exponents from the first

identify equivalent numbers 24 and

two tables and then use the third

26, from which

table to compute the logarithm.

16 ? 64 24 ? 26 246 210 1024.

Part of a page from Napier's Logarithmic Tables.

After his logarithms of numbers, Napier produced a table to give seven-place logarithms of

To be useful, of course, tables must identify the exponents of all the numbers we

sines of angles for every minute from 0 to 90. Kepler credited Napier's tables for

might need to multiply.

making possible the incredible calculations

John Napier (1550?1617) spent twenty years required to analyze the motion of the planets

compiling tables of exponents (called logarithms

about the sun.

or ratio numbers). He started with a large number

Strategy: Rewrite each part as needed to use inverse function identities.

EXAMPLE 3 Using inverse function identities Use inverse function identities to simplify. Express the result in exact form and then give a five-decimalplace approximation.

(a) e ln 3

(b) e2 ln 7

(c) ln e5

Solution

(a) By identity (1), eln 3 3 1.73205. The exact form is 3 and 1.73205 is the desired approximation.

(b)For the exact form, first use logarithm property L3 to rewrite 2 ln 7 as ln

72,

or

ln

1 49

,

then

use

identity

(1).

e2 ln 7 e ln (149) 1 0.02041 49

4.4 Computations with Logarithmic and Exponential Functions

231

y y = e lnx x

(0, 0)

(a) y

y = ln e x x

(0, 0)

(b) FIGURE 23

Strategy: Apply the natural logarithm function to both sides and simplify, using properties of logarithms.

Thus

e2

ln

7

is

exactly

equal

to

1 49

and

0.02041

is

the

five-decimal-place

approx-

imation.

(c) Identity (2) gives ln e5 5 2.23607. An exact form for ln e5 is

5 and 2.23607 is the desired approximation.

EXAMPLE 4 Identical functions? Graph the functions f x eln x and gx ln ex separately. Describe and explain the differences between the graphs of f , g, and the line y x.

Solution The graphs of y eln x and y ln ex are shown in Figures 23a and 23b. The graph of f is the same as the first quadrant portion of the line y x, but the domain of f is limited to x 0. We cannot tell visually whether the origin is included, but in a decimal window, tracing verifies that f is undefined at x 0.

Since ex is always positive, ln ex x is defined for all real numbers x. The graph of g is identical with the graph of y x.

The graphs of both f and g give graphical confirmation of the inverse function identities.

Using Inverse Function Identities to Solve Equations

In Section 4.1 we solved the equation 32x1 373 by using our intuitive under-

standing of exponents. To justify equating exponents, we now know that exponen-

tial and logarithmic functions are one?one; if two numbers are equal, their loga-

rithms are equal, or in mathematical notation, if u v, then logb u logb v.

Applying the log function to both sides, if 32x1 373, then log332x1

log3373,

from

which

2x

1

7 3

,

and

so

x

2 3

.

EXAMPLE 5 Solving exponential equations Solve. Express your solution in exact form and give a four-decimal-place approximation.

(a) e2x1 4 (b) 5x 3 ? 41x

Solution (a) From the strategy,

ln e2x1 ln 4 or 2x 1 ln 4.

Therefore x

1

ln 2

4

1.1931, so

1

ln 2

4

is

the

exact

solution

and

1.1931 is

the

desired approximation.

(b) In a similar fashion, ln 5x ln3 ? 41x. By logarithm property L3, ln 5x

x ln 5, and by properties L1 and L3, ln3 ? 41x ln 3 1 xln 4. There-

fore, the given equation is equivalent to

x ln 5 ln 3 1 xln 4.

We now have a linear equation in x. Solve it as follows:

x ln 5 ln 3 ln 4 x ln 4

xln 5 ln 4 ln 3 ln 4

x

ln ln

3 5

ln ln

4 4

L1

ln ln

12 20

0.8295.

Therefore,

the

exact

solution

is

ln ln

12 20

and

0.8295

is

the

approximation.

232

Chapter 4 Exponential and Logarithmic Functions

Notice

that

ln 12 ln 20

cannot

be

simplified

further

in

the

exact

form

solution

of

Example

5.

In

particular,

ln 12 ln 20

is

not

equal

to

ln

, 12

20

since

ln 12 ln 20

0.8295

and

ln

12 20

0.5108.

EXAMPLE 6 Another exponential equation Solve the equation ex ex 4.

Strategy: Note that the

strategy of Example 5 is not helpful, since lnex ex does not simplify. Multiply through by ex to get a quadratic equation in ex. Use

the quadratic formula to solve for ex, and then take

logarithms to solve for x.

B (1, 3) y=x

A (0, 1)

D

y = 3x

C

(3, 1)

(1, 0) y = log3 x

Solution Follow the strategy and multiply both sides by ex.

e2x exex 4ex or ex2 4ex 1 0.

Use the quadratic formula to solve for ex,

ex 2 3 and ex 2 3.

Apply the ln function to both sides of each and use identity (2) to get

ln ex ln2 3 or x ln2 3 1.317 ln ex ln2 3 or x ln2 3 1.317.

The exact solutions are ln2 3 and ln2 3. Decimal approximations are 1.317 and 1.317, respectively.

EXAMPLE 7 Inverse functions Graph the functions f x 3x, gx log3 x, and y x in the same decimal window. Find at least two pairs of points on the graphs of f and g that are reflections of each other in the line y x. What do the graphs suggest about the domain and range of f and g?

[? 5, 5] by [? 3.5, 3.5] FIGURE 24

Solution To graph y log3 x, we use the change of base formula and enter Y LN X/LN 3. The graphs of all three are shown in Figure 24.

From the figure, it looks as if the graph of g is the reflection of the graph of f in the line y x. For partial verification, we trace along the graph of y 3x and find points A0, 1 and B1, 3. On the graph of y log3 x are the points C1, 0, the reflection of point A, and D3, 1, the image of B.

Exponential functions are said to "grow faster" than any polynomial function. We are not prepared to prove such a general statement, but it is illustrative to see how unexpected the intersections of polynomial and exponential functions may be, as suggested in the next example.

EXAMPLE 8 Hidden intersections Let f x 2x and gx x 3.

(a) Graph f and g in the 1, 6 1, 10 window. To the right of the visible intersection, which graph appears to be growing faster? Use an x-range of 1, 12 and keep increasing the y-range until you find another intersection.

(b) Find the "hidden intersection," (one decimal place) by setting 2x x 3 and taking the natural logarithm of both sides.

(c) Discuss alternative ways to use technology to find the intersection in part (b).

4.4 Computations with Logarithmic and Exponential Functions

233

y = 3ln x

Q

y = x ln 2

P

[? 1, 10] by [? 1, 7] FIGURE 25

Solution

(a) In the specified window, there is an intersection near (1.4, 2.7), and then the cubic function rises much faster than the exponential. Trying larger and larger y-ranges, it isn't until we get to something near 1000 that the exponential function "catches up." Tracing, the intersection is close to (9.9, 980). From this point on, the exponential graph grows faster.

(b) When we take the natural logarithm of both sides of the equation 2x x 3 and apply properties of logarithms, we get the equation, x ln 2 3 ln x, for which we have no direct way of solving. Nevertheless, graphical tools are available. Graphing Y1 X LN 2 and Y2 3 LN X in 1, 10 1, 7 gives a picture something like Figure 25. The intersection Q is near (9.94, 6.89). The "hidden intersection" of the original graphs is given by x 9.94, for which y 29.94 982 and 9.943 982.

(c) Among the many alternative approaches using a graphing calculator, we could locate graphically the root of either the equation 2x x 3 0 or of x ln 2 3 ln x 0, or, if our calculator has one, we could use a solve routine (mentioned in Example 10 of Section 4.1) for any of the above equations. Any of the solve routines require a starting guess. In this case, we must indicate that we want the solution near 9.9, which we will find is approximately 9.9395351414. In summary, we conclude that the equation 2x x 3 has two roots, x1 1.4 and x2 9.9.

Applications

Exponential and logarithmic functions are used to model many natural phenomena. The following section is devoted entirely to such applications. Here we discuss just one example.

The sounds we hear Logarithmic functions are used in modeling the sounds we

hear. Loudness of sound is a sensation in the brain. We cannot measure it directly,

but there is a related physically measurable quantity: the intensity of the sound

wave. Sound waves travel through the air, and these wave vibrations force the

eardrums to vibrate, producing a sound sensation. The intensity I of a sound wave

is

measured

in

watts

per

square

meter

. w m2

The

intensity

of

a

barely

audible

sound

wave,

about

1012

w m2

,

corresponds

to

pressure vibrations less than a billionth of the atmospheric pressure at sea level.

The

human

ear

is

very

sensitive.

A

sound

wave

of

intensity

of

1

w m2

would

damage

the eardrum.

The human ear does not respond to sound intensity in a linear fashion. If the

intensity doubles, we do not hear the sound as twice as loud. The sound level is

logarithmically related to the intensity I.

I 10 log I I0

10log I log I0

(3)

where I is the measured intensity and I0 is the intensity of sound we can just barely

hear,

10 12

w m2

.

The

sound

level

is

measured

in

decibels

(dB),

a

unit

named

for

Alexander Graham Bell.

234

Chapter 4 Exponential and Logarithmic Functions

For a sound just at the hearing threshold, I is I0, so

I 10 log I0 10 log 1 10 ? 0 0. I0

Thus 0 dB measures the threshold hearing level. At an intensity of 10 I0, 10 I0 10 log 10 10. Similarly, if I is 100 I0, then the sound level is given by 100 I0 10 log 100 10 ? 2 20. Multiplying the intensity by a factor of 10 only doubles the loudness of the sound we hear.

EXAMPLE 9 Adding trumpets Four trumpets are playing at the same time, each at an average loudness of 75 dB. What is the resulting sound level?

Solution

If I1 denotes the loudness level of one trumpet, then Equation (3) can give the

corresponding intensity I1.

I1 10 log

I1 I0

10 log I1 10 log I0.

Since I0 1012, log I0 12. Since I1 75, we have

75 10 log I1 120 log I1 4.5 and I1 10 . 4.5

The intensity of sound for one trumpet is 104.5 so four trumpets have a sound

intensity of 4 ? I1, or 4 ? 10 . 4.5 Thus

4 ? I1 10 log

4 I1 I0

10 log 4 I1 I0

10 log 4 10 log I1 10 log 4 75 81.02

I0

Therefore, the loudness of the four trumpets is about 81 dB. A fourfold increase in sound wave intensity increases the loudness level by less than 10 percent. This is why a solo instrument can be heard in a symphony concert even when the full orchestra is playing at the same time.

EXERCISES 4.4

Check Your Understanding

Exercises 1?5 True or False. Give reasons. 1. log 16 ln 5 2. ln2 5 1 ln 2 ln 5

2 3. For all positive numbers c and d,

lnc d ln c ln d. 4. The graph of y log x is above the graph of y ln x

for all x 1. 5. The graph of y ln x is above the graph of y log3 x

for every x 1.

Exercises 6?10 Fill in the blank so that the resulting statement is true.

6. The number of integers between ln 4 and 5ln 25 is .

7. The sum of the integers between ln 4 and 2ln 25 is .

8. If S x ln 0.5 x 5ln 25, then the smallest

positive integer that is not in S is

.

9. The graph of y lnx 3 x 2 4x 6 has turning

points in Quadrants

.

10. The local minimum point (2 decimal places) for the

graph of y 2 lnx 3 x 2 4x 4 is

.

Develop Mastery

Exercises 1?8 Logarithmic Evaluations Evaluate. Give the result rounded off to four decimal places. If your calculator indicates an error, explain why.

1. (a) ln 5 (b) log 15.6 2. (a) ln 3 (b) log1 3

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