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A. Solving Logarithmic and Exponential Equations

A. Solving Logarithmic and Exponential Equations

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EXAMPLE 1







5–50



CHAPTER 5 Exponential and Logarithmic Functions







Solving a Logarithmic Equation



Solve for x and check your answer: log x ϩ log 1x ϩ 32 ϭ 1.





Algebraic Solution



log x ϩ log 1x ϩ 32 ϭ 1

log 3x 1x ϩ 32 4 ϭ 1

x2 ϩ 3x ϭ 101

x2 ϩ 3x Ϫ 10 ϭ 0

1x ϩ 52 1x Ϫ 22 ϭ 0

x ϭ Ϫ5 or x ϭ 2



original equation

product property

exponential form,

distribute x

set equal to 0

factor

result



Graphical Solution



Using the intersection-ofgraphs method, we enter

Y1 ϭ log X ϩ log1X ϩ 32

and Y2 ϭ 1. From the domain

we know x 7 0, indicating

the solution will occur in QI.

After graphing both functions

using the window shown, the

intersection method shows

the only solution is x ϭ 2.



3



0



5



Ϫ3



Check: The “solution” x ϭ Ϫ5 is outside the domain and is ignored. For x ϭ 2,

log x ϩ log1x ϩ 32 ϭ 1 original equation

log 2 ϩ log12 ϩ 32 ϭ 1 substitute 2 for x

log 2 ϩ log 5 ϭ 1 simplify

log12 # 52 ϭ 1 product property

log 10 ϭ 1 Property I

You could also use a calculator to verify log 2 ϩ log 5 ϭ 1 directly.

Now try Exercises 7 through 14







If the simplified form of an equation yields a logarithmic term on both sides, the

uniqueness property of logarithms provides an efficient way to work toward a solution. Since logarithmic functions are one-to-one, we have

The Uniqueness Property of Logarithms

For positive real numbers m, n, and b 1,

1. If logb m ϭ logb n,

2. If

m

then logbm

then m ϭ n

Equal bases imply equal arguments.



EXAMPLE 2







n,

logbn



Solving Logarithmic Equations Using the Uniqueness Property

Solve each equation using the uniqueness property.

a. log 1x ϩ 22 ϭ log 7 ϩ log x

b. ln 87 Ϫ ln x ϭ ln 29











Algebraic Solution



a. log 1x ϩ 22

log 1x ϩ 22

xϩ2

2

1

3



ϭ log 7 ϩ log x

ϭ log 7x

ϭ 7x

ϭ 6x

ϭx



properties of logarithms

uniqueness property

solve for x

result



Graphical Solution



Deciding which method to use (intersection or zeroes)

can depend on the simplicity or complexity of the

equation, how the equation is given, and/or which

method gives the clearest view of the point(s) of

intersection. Here, we opt to use the zeroes method.



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Section 5.5 Solving Exponential and Logarithmic Equations



b. ln 87 Ϫ ln x ϭ ln 29

87

ln a b ϭ ln 29

x

87

ϭ 29

x

87 ϭ 29x

3ϭx



Figure 5.44

a. After setting the

2

equation equal to

zero, we enter

Y1 ϭ log1X ϩ 22 Ϫ

log 7 Ϫ log X

0

5

and locate the

x-intercept (if it

exists) using 2nd

TRACE (CALC)

Ϫ2

2:Zero. The graph

reveals an

x-intercept between 0 and 1, and using these values as

1

bounds we locate the zero at x ϭ (Figure 5.44).

3

b. Once again

Figure 5.45

setting the

2

equation equal to

zero, we enter

Y1 ϭ ln 87 Ϫ

ln X Ϫ ln 29 to

0

10

locate the

x-intercept (if it

exists). The graph

shows an

Ϫ2

x-intercept near 3, with the calculator indicating the

zero is exactly x ϭ 3 (Figure 5.45).



quotient property



uniqueness property

clear denominator

result



Now try Exercises 15 through 20



WORTHY OF NOTE

The uniqueness property can also

be viewed as exponentiating both

sides using the appropriate base,

then applying Property IV.



EXAMPLE 3







Often the solution may depend on using a variety of algebraic skills in addition to

logarithmic or exponential properties.







Solving Logarithmic Equations

Solve the equation and check your answer.

log 1x ϩ 122 Ϫ log x ϭ log 1x ϩ 92







529







Algebraic Solution



log 1x ϩ 122 Ϫ log x ϭ log 1x ϩ 92

x ϩ 12

b ϭ log 1x ϩ 92

log a

x

x ϩ 12

ϭxϩ9

x

x ϩ 12 ϭ x2 ϩ 9x

0 ϭ x2 ϩ 8x Ϫ 12



given equation

quotient property



uniqueness property

clear denominator

set equal to 0



Graphical Solution



Using the intersection-of-graphs method, we

enter log 1X ϩ 122 Ϫ log x as Y1 and

log1X ϩ 92 as Y2 on the Y= screen. Using

TRACE (CALC) 5:intersect, we find the

2nd

graphs intersect at x ϭ 1.2915026, and that

this is the only solution (knowing the graphs’

basic shape, we conclude they cannot intersect

again).



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CHAPTER 5 Exponential and Logarithmic Functions



The equation is not factorable, and the quadratic

formula must be used.



ϭ



Ϫb Ϯ 2b2 Ϫ 4ac

2a

Ϫ8 Ϯ 2182 2 Ϫ 41121Ϫ122



2112

Ϫ8 Ϯ 1112

Ϫ8 Ϯ 4 17

ϭ

ϭ

2

2

ϭ Ϫ4 Ϯ 2 17



5



quadratic formula



Ϫ10



10



substitute 1 for a, 8

for b, Ϫ12 for c

Ϫ5



simplify

result



Substitution shows x ϭ Ϫ4 ϩ 2 17 1x Ϸ 1.291502 checks, but substituting

Ϫ4 Ϫ 2 17 for x gives log 12.70852 Ϫ log 1Ϫ9.29152 ϭ log 1Ϫ0.29152 and two of

the three terms do not represent real numbers (x ϭ Ϫ4 Ϫ 217 is an extraneous root).

Now try Exercises 21 through 36







CAUTION







Be careful not to dismiss or discard a possible solution simply because it’s negative. For

the equation log1Ϫ6 Ϫ x2 ϭ 1, x ϭ Ϫ16 is the solution (the domain here allows negative

numbers: Ϫ6 Ϫ x 7 0 yields x 6 Ϫ6 as the domain). In general, when a logarithmic

equation has multiple solutions, all solutions should be checked.



Solving an exponential equation likewise involves isolating an exponential term

on one side, or writing the equation where exponential terms of like base occur on each

side. The latter case can be solved using the uniqueness property. If the exponential

base is neither 10 nor e, logarithms of base b can be used along with the change-ofbase formula to solve the equation.



EXAMPLE 4







Solving an Exponential Equation Using Base b

Solve the exponential equation. Answer in both exact form, and approximate form

to four decimal places: 43x Ϫ 1 ϭ 8



Solution







43x Ϫ 1 ϭ 8

43x ϭ 9



given equation

add 1



The left-hand side is neither base 10 or base e, so here we chose base 4 to solve.

log443x ϭ log49

log 9

3x ϭ

log 4

log 9



3 log 4

x Ϸ 0.5283



logarithms base 4

Property III; change-of-base property

multiply by 13 (exact form)

approximate form



A calculator check is shown here.

Now try Exercises 37 through 40







In some cases, two exponential terms with unlike bases may be involved. In

this case, either common logs or natural logs can be used, but be sure to distinguish

between constant terms like ln 5 and variable terms like x ln 5. As with all equations, the goal is to isolate the variable terms on one side.



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Section 5.5 Solving Exponential and Logarithmic Equations



EXAMPLE 5







Solving an Exponential Equation with Unlike Bases

Solve the exponential equation 5xϩ1 ϭ 62x.



Algebraic Solution







5xϩ1 ϭ 62x



original equation



Begin by taking the natural log of both sides:

ln 15xϩ1 2 ϭ ln 162x 2

1x ϩ 12 ln 5 ϭ 2x ln 6

x ln 5 ϩ ln 5 ϭ 2x ln 6

ln 5 ϭ 2x ln 6 Ϫ x ln 5

ln 5 ϭ x12 ln 6 Ϫ ln 52

ln 5

ϭx

2 ln 6 Ϫ ln 5

0.8153 Ϸ x



Graphical Solution







apply base-e logarithms

power property

distribute

variable terms to one side

factor out x

solve for x (exact form)

approximate form



In many cases, the quality of a graphical solution depends on the ability to

determine an appropriate window size. Most often, this is accomplished using the

domain of the functions involved, the context of the application, or a few test

values. For Y1 ϭ 5Xϩ1 and Y2 ϭ 62X, we begin by observing that for x ϭ 0,

Y1 7 Y2. However for x ϭ 2, Y2 7 Y1 (see Figure 5.46). This indicates that

function values will be equal 3 Y1 1X2 ϭ Y2 1X2 4 for some value of x between 0 and

1, and the window size for x must be set accordingly. These test values also show

that y need not be greater than 36, though we may elect to use a higher value for

both x and y to obtain a good window “frame.” Using the window size indicated

in Figure 5.47 reveals the solution to the equation (where the graphs intersect) is

x Ϸ 0.81528463.

Figure 5.47



Figure 5.46



50



0



1.5



Ϫ15



Now try Exercises 41 through 44







As an alternative to taking the natural log of both sides directly, we can use the

properties of exponents to simplify and combine the exponential terms as follows.

5xϩ1 ϭ 62x



original equation



5 5 ϭ 16 2

x 1



2 x



product and power properties



5 # 5x ϭ 36x



x



rewrite factors; simplify

x



36

36

ϭa b

5x

5

ln 5 ϭ x ln 7.2

ln 5

ϭx

ln 7.2





divide by 5x, apply power property

take natural logs 1 36

5 ϭ 7.22

solve for x



The result is equivalent to our original solution: x Ϸ 0.81528463.



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CHAPTER 5 Exponential and Logarithmic Functions



Logarithmic equations come in many different forms, and the following ideas

summarize the basic approaches used in solving them. For this summary, recall that for

any positive real number k, logbk is a constant. Assume M, N, and X represent algebraic

expressions in x.

1. If the equation can be written in the form logb M ϭ constant, use the exponential

form and algebra to solve: bconstant ϭ M.

2. If the equation can be written in the form logb M ϭ logb N, use the uniqueness

property and algebra to solve: M ϭ N.

3. If the equation has an additional constant term as in logb M ϭ logb N ϩ constant,

move all logarithmic terms to one side and consolidate using the product or quotient properties, then use the exponential form and algebra to solve:

logb M ϭ logb N ϩ constant

logb M Ϫ logb N ϭ constant

logb a



M

b ϭ constant

N

M

ϭ bconstant

N



4. If the equation has multiple logarithmic terms as in logb X ϭ logb M ϩ logb N,

consolidate logarithmic terms using the product or quotient properties, then use

the uniqueness property and algebra to solve:

logb X ϭ logb M ϩ logb N



logb X ϭ logb 1MN2

X ϭ MN



Many other forms and varieties are possible.

In advanced applications, the equations used are sometimes impossible to solve

using inverse functions. This is often the case when logarithmic or exponential functions are mixed with other functions (polynomial, radical, rational, etc.). In these cases

graphing and calculating technologies become indispensible tools, and the emphasis in

working towards a solution shifts more to an understanding of the domain, the graphical attributes of the functions involved, and setting an appropriate window.

EXAMPLE 6







Solving Equations Using Technology

3



Solution







Find all solutions to e1xϪ8 ϭ 1x ϩ 9.

Begin by noting that the domain of the

function on the left, call it Y1, is all real

numbers, since this is the domain of both

3

y ϭ ex and y ϭ 1

x. However, the domain of

the function on the right (call it Y2) is

x Ն Ϫ9, and we expect that any solution(s)

to the equation must occur to the right of

Ϫ9, so we opt for a standard viewing

window to begin (Figure 5.48). At first it

appears the graphs do not intersect to the

left, but our knowledge of the domain



Figure 5.48

10



Ϫ10



10



Ϫ10



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Section 5.5 Solving Exponential and Logarithmic Equations



A. You’ve just seen how we

can solve general logarithmic

and exponential equations



indicates they must intersect near x ϭ Ϫ9,

since Y1 and Y2 are both positive. Using the

intersection-of-graphs method, we find one

solution is x Ϸ Ϫ8.994. To the right, no

point of intersection is initially visible so

we extend our window to explore whether

the graphs intersect again. Using Xmax ϭ 20

we note the graphs indeed intersect

(Figure 5.49) and locate a second solution

at x Ϸ 11.433.



533



Figure 5.49

10



Ϫ10



20



Ϫ10



Now try Exercises 45 through 48







B. Applications of Logistic, Exponential, and

Logarithmic Functions

Applications of exponential and logarithmic functions take many different forms and it

would be impossible to illustrate them all. As you work through the exercises, try to adopt

a “big picture” approach, applying the general principles illustrated here to other applications. Some may have been introduced in previous sections. The difference here is that we

can now solve for the independent variable, instead of simply evaluating the relationships.

In applications involving the logistic growth of animal populations, the initial

stage of growth is virtually exponential, but due to limitations on food, space, or other

resources, growth slows and at some point it reaches a limit. In business, the same principle applies to the logistic growth of sales or profits, due to market saturation. In these

cases, the exponential term appears in the denominator of a quotient, and we “clear

denominators” to begin the solution process.

EXAMPLE 7







Solving a Logistic Equation

A small business makes a new discovery and begins an aggressive advertising

campaign, confident they can capture 66% of the market in a short period of time.

They anticipate their market share will be modeled by the function

66

M1t2 ϭ

, where M(t) represents the percentage after t days. Use this

1 ϩ 10eϪ0.05t

function to answer the following.

a. What was the company’s initial market share (t ϭ 0)? What was their market

share 30 days later?

b. How long will it take the company to reach a 60% market share?











Algebraic Solution



a. M1t2 ϭ

M102 ϭ



66

1 ϩ 10eϪ0.05t

66



1 ϩ 10eϪ0.05102

66

ϭ

11

ϭ6



given



substitute 0 for t

simplify 1e 0 ϭ 12

result



Graphical Solution



a. After entering the function as

Y1, we can find Y1(0) and

Y1(30) directly on the home

screen (Figure 5.50). The

company began the ad

campaign with a 6% market

share, and 30 days later they

had secured over a 20.4%

market share.



Figure 5.50



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The company originally had only a 6% market share.

M1302 ϭ



66



substitute 30 for t



1 ϩ 10eϪ0.051302

66

ϭ

1 ϩ 10eϪ1.5

Ϸ 20.4



simplify

result



After 30 days, they held a 20.4% market share.

b. For part (b), we replace M(t) with 60 and

solve for t.

66

1 ϩ 10eϪ0.05t

6011 ϩ 10eϪ0.05t 2 ϭ 66

1 ϩ 10eϪ0.05t ϭ 1.1

10eϪ0.05t ϭ 0.1

eϪ0.05t ϭ 0.01

ln eϪ0.05t ϭ ln 0.01

Ϫ0.05t ϭ ln 0.01

ln 0.01



Ϫ0.05

Ϸ 92

60 ϭ



given

multiply by 1 ϩ 10eϪ0.05t

divide by 60

subtract 1

divide by 10



b. Using the intersection-of-graphs method, we

graph Y1 with Y2 ϭ 60 to find any point(s) of

intersection. For the window size, we reason

that after 30 days, there is only a 20.4% market

share (x must be much greater than 30), and a

60% market share is being explored (y must be

greater than 60). Using the window indicated in

Figure 5.51 reveals that a 60% market share

will be attained shortly after the 92nd day.



apply base-e logarithms



Figure 5.51



Property III



80



solve for t (exact form)

approximate form

0



120



0



The company will reach a 60% market share in about 92 days.

Now try Exercises 49 and 50







P0

b to find an

P

altitude H, given a temperature and the atmospheric (barometric) pressure in centimeters of mercury (cmHg). Using the tools from this section, we are now able to find the

atmospheric pressure for a given altitude and temperature.

Earlier we used the barometric equation H ϭ 130T ϩ 80002 ln a



EXAMPLE 8







Using Logarithms to Determine Atmospheric Pressure

Suppose a group of climbers has just scaled Mt. Rainier, the highest mountain of

the Cascade Range in western Washington State. If the mountain is about 4395 m

high and the temperature at the summit is Ϫ22.5°C, what is the atmospheric

pressure at this altitude? The pressure at sea level is P0 ϭ 76 cmHg.



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