A. Solving Logarithmic and Exponential Equations
Tải bản đầy đủ - 0trang
cob19545_ch05_528-538.qxd
9/2/10
9:48 PM
Page 528
College Algebra Graphs & Models—
528
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.
cob19545_ch05_528-538.qxd
9/3/10
8:29 PM
Page 529
College Algebra Graphs & Models—
5–51
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).
cob19545_ch05_528-538.qxd
9/2/10
9:48 PM
Page 530
College Algebra Graphs & Models—
530
5–52
CHAPTER 5 Exponential and Logarithmic Functions
The equation is not factorable, and the quadratic
formula must be used.
xϭ
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
xϭ
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.
cob19545_ch05_528-538.qxd
1/7/11
5:06 PM
Page 531
College Algebra Graphs & Models—
5–53
531
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
5ϭ
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.
cob19545_ch05_528-538.qxd
9/2/10
9:48 PM
Page 532
College Algebra Graphs & Models—
532
5–54
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
cob19545_ch05_528-538.qxd
11/27/10
12:39 AM
Page 533
College Algebra Graphs & Models—
5–55
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
cob19545_ch05_528-538.qxd
9/2/10
9:48 PM
Page 534
College Algebra Graphs & Models—
534
5–56
CHAPTER 5 Exponential and Logarithmic Functions
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
tϭ
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.