Download SECTION 4.6 4.6 Logarithmic and Exponential Equations

SECTION 4.6
4.6
Logarithmic and Exponential Equations
309
Logarithmic and Exponential Equations
PREPARING FOR THIS SECTION
Before getting started, review the following:
• Solving Equations Using a Graphing Utility (Section 1.3, pp. 24–26)
• Solving Quadratic Equations (Appendix, Section A.5,
pp. 988–996)
Now work the ‘Are You Prepared?’ problems on page 313.
OBJECTIVES
1
2
3
Solve Logarithmic Equations Using the Properties of Logarithms
Solve Exponential Equations
Solve Logarithmic and Exponential Equations Using a Graphing Utility
1 Solve Logarithmic Equations Using the Properties
✓
of Logarithms
In Section 4.4 we solved logarithmic equations by changing a logarithm to exponential form. Often, however, some manipulation of the equation (usually using the
properties of logarithms) is required before we can change to exponential form.
Our practice will be to solve equations, whenever possible, by finding exact solutions using algebraic methods and exact or approximate solutions using a graphing utility. When algebraic methods cannot be used, approximate solutions will be
obtained using a graphing utility. The reader is encouraged to pay particular attention to the form of equations for which exact solutions are possible.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
310
Exponential and Logarithmic Functions
CHAPTER 4
EXAMPLE 1
Solving a Logarithmic Equation
Solve:
2 log 5 x = log5 9
Algebraic Solution
Graphing Solution
Because each logarithm is to the same base, 5, we can obtain an exact
solution as follows:
To solve the equation using a graphing
2 log x
utility, graph Y1 = 2 log5 x =
and
log 5
2 log 5 x
log 5 x2
x2
x = 3 or
✔ CHECK:
=
=
=
x
2 log 5 3 log 5 9
log 5 32 log 5 9
log5 9 = log 5 9
log 5 9
log 5 9
9
= -3
loga Mr = r loga M
If loga M = loga N, then M = N.
Recall that logarithms of negative numbers are not defined, so,
in the expression 2 log5 x, x must
be positive. Therefore, -3 is extraneous and we discard it.
log 9
, and determine the point
log 5
of intersection. See Figure 45.
Y2 = log 5 9 =
Figure 45
4
4
r loga M = loga M r
6
The solution set is 536.
4
NOW WORK PROBLEM
EXAMPLE 2
9.
Solving a Logarithmic Equation
Solve:
log 41x + 32 + log 412 - x2 = 1
Algebraic Solution
Graphing Solution
To obtain an exact solution, we need to express the left side as a single
logarithm. Then we will change the expression to exponential form.
Graph Y1 = log 41x + 32 + log 412 - x2 =
log41x + 32 + log 412 - x2 = 1
log 431x + 3212 - x24 = 1
1x + 3212 - x2 = 4 = 4
2
-x - x + 6 = 4
x2 + x - 2 = 0
1x + 221x - 12 = 0
x = -2 or x = 1
1
loga M + loga N = loga(MN)
Change to an exponential
expression.
log1x + 32
log12 - x2
and Y2 = 1 and
+
log 4
log 4
determine the points of intersection. See
Figure 46.
Figure 46
2
Simplify.
Place the quadratic equation in
standard form.
3
Factor.
Zero-Product Property
3
1
Since the arguments of each logarithmic expression in the equation are
positive for both x = -2 and x = 1, neither is extraneous. We leave the
check to you.
2
3
The solution set is 5-2, 16.
NOW WORK PROBLEM
3
1
13.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.6
Logarithmic and Exponential Equations
311
2 Solve Exponential Equations
✓
In Sections 4.3 and 4.4, we solved certain exponential equations algebraically by expressing each side of the equation with the same base. However, many exponential equations
cannot be rewritten so each side has the same base. In such cases, properties of logarithms along with algebraic techniques can sometimes be used to obtain a solution.
EXAMPLE 3
Solving an Exponential Equation
Solve:
4 x - 2 x - 12 = 0
Algebraic Solution
We note that 4 = 12 2 = 2 = 12 2 , so the equation is actually quadratic in form, and we can rewrite it as
2 x
x
2x
12 x2 - 2 x - 12 = 0
2
x 2
Graphing Solution
Graph Y1 = 4 x - 2 x - 12 and determine
the x-intercept. See Figure 47.
Figure 47
Let u = 2x; then u2 - u - 12 = 0.
100
Now we can factor as usual.
12 x - 4212 x + 32 = 0
(u - 4)(u + 3) = 0
2 - 4 = 0 or
2x + 3 = 0
u - 4 = 0 or u + 3 = 0
x
x
2 = 4
2 = -3
u = 2x = 4
u = 2x = -3
x
The equation on the left has the solution x = 2, since 2 = 4 = 2 2; the
equation on the right has no solution, since 2 x 7 0 for all x. The only solution is 2.
The solution set is 526.
x
1
4
30
In the algebraic solution of the previous example, we were able to write the exponential expression using the same base after utilizing some algebra, obtaining an
exact solution to the equation. When this is not possible, logarithms can sometimes
be used to obtain the solution.
EXAMPLE 4
Solving an Exponential Equation
Solve:
2x = 5
Algebraic Solution
Graphing Solution
Since 5 cannot be written as an integral power of 2, we write the exponential
equation as the equivalent logarithmic equation.
2x = 5
ln 5
x = log 2 5 =
ln 2
q
Graph Y1 = 2 x and Y2 = 5 and determine the x-coordinate of the
point of intersection. See Figure 48.
Figure 48
16
Change-of-Base Formula (9), Section 4.5
Alternatively, we can solve the equation 2 x = 5 by taking the natural logarithm
(or common logarithm) of each side. Taking the natural logarithm,
2x = 5
ln 2 x = ln 5
If M = N, then ln M = ln N.
x ln 2 = ln 5
ln Mr = r ln M
ln 5
Exact solution
x =
ln 2
Approximate solution
L 2.322
NOW WORK PROBLEM
1
4
4
The approximate solution, rounded
to three decimal places, is 2.322. 21.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
312
Exponential and Logarithmic Functions
CHAPTER 4
EXAMPLE 5
Solving an Exponential Equation
Solve:
8 # 3x = 5
Algebraic Solution
Graphing Solution
8 # 3x = 5
3x =
5
8
Graph Y1 = 8 # 3x and Y2 = 5 and determine the xcoordinate of the point of intersection. See Figure 49.
Solve for 3x.
5
lna b
5
8
x = log 3 a b =
8
ln 3
Figure 49
24
Exact solution
2
L -0.428
Approximate solution
1
8
The approximate solution, rounded to three decimal
places, is –0.428.
EXAMPLE 6
Solving an Exponential Equation
Solve:
5x - 2 = 33x + 2
Algebraic Solution
Graphing Solution
Because the bases are different, we first apply Property (6), Section 4.5
(take the natural logarithm of each side), and then use appropriate properties of logarithms. The result is an equation in x that we can solve.
Graph Y1 = 5x - 2 and Y2 = 33x + 2 and
determine the x-coordinate of the point of
intersection. See Figure 50.
5x - 2
ln 5x - 2
1x - 22 ln 5
1ln 52x - 2 ln 5
1ln 52x - 13 ln 32x
1ln 5 - 3 ln 32x
33x + 2
ln 33x + 2
13x + 22 ln 3
13 ln 32x + 2 ln 3
2 ln 3 + 2 ln 5
21ln 3 + ln 52
21ln 3 + ln 52
x =
ln 5 - 3 ln 3
L -3.212
=
=
=
=
=
=
Figure 50
If M = N, ln M = ln N.
0.0015
ln Mr = r ln M
Distribute.
Place terms involving x on the left.
Factor.
4
Exact solution
Approximate solution
0
0.0005
The approximate solution, rounded to three
decimal places, is –3.212.
NOW WORK PROBLEM
29.
3 Solve Logarithmic and Exponential Equations Using a
✓
Graphing Utility
The algebraic techniques introduced in this section to obtain exact solutions apply
only to certain types of logarithmic and exponential equations. Solutions for other
types are usually studied in calculus, using numerical methods. For such types, we
can use a graphing utility to approximate the solution.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall
SECTION 4.6
EXAMPLE 7
Figure 51
Logarithmic and Exponential Equations
313
Solving Equations Using a Graphing Utility
Solve: x + ex = 2
Express the solution(s) rounded to two decimal places.
4
Y1 x e x
The solution is found by graphing Y1 = x + ex and Y2 = 2. Y1 is an increasing function (do you know why?), and so there is only one point of intersection
for Y1 and Y2 . Figure 51 shows the graphs of Y1 and Y2 . Using the INTERSECT
command, the solution is 0.44 rounded to two decimal places.
Solution
Y2 2
0
1
0
NOW WORK PROBLEM
53.
Copyright © 2006 Pearson Education, Inc., publishing as Pearson Prentice Hall