Download Solving Exponential Equations

Section 5.5
Solving Exponential
and
Logarithmic
Equations
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.
Objectives


Solve exponential equations.
Solve logarithmic equations.
Solving Exponential Equations
Equations with variables in the exponents, such as
3x = 20
and
25x = 64,
are called exponential equations.
Use the following property to solve exponential
equations.
Base-Exponent Property
For any a > 0, a  1,
ax = ay  x = y.
Example
3x7
 32.
Solve: 2
Write each side as a power of the same number (base).
2 3x7  2 5
Since the bases are the same number, 2, we can use
the base-exponent property and set the exponents
equal:
3x7
2
 32.
Check x = 4:
3x  7  5
34 7
2
? 32
3x  12
127
2
x4
25
The solution is 4.
32 32 TRUE
Another Property
Property of Logarithmic Equality
For any M > 0, N > 0, a > 0, and a  1,
loga M = loga N  M = N.
Example
Solve: 3x = 20.
3x  20
log 3  log 20
x
x log 3  log 20
This is an exact answer. We
cannot simplify further, but we
can approximate using a
calculator.
log 20
x
 2.7268
log 3
We can check by finding 32.7268  20.
Example
Solve: 100e0.08t = 2500.
100 e0.08t  2500
ln e0.08t  ln 25
0.08t  ln 25
ln 25
t
0.08
t  40.2
The solution is about 40.2.
Example
Solve: 4x+3 = 3-x.
4 x 3  3 x
log 4
x 3
x
 log3
( x  3)log 4   x log3
x log 4  x log 3  3log 4
x(log 4  log3)  3log 4
3log 4
x
log 4  log 3
x  1.6737
Solving Logarithmic Equations
Equations containing variables in logarithmic
expressions, such as
log2 x = 4
and
log x + log (x + 3) = 1,
are called logarithmic equations.
To solve logarithmic equations algebraically, we first try to
obtain a single logarithmic expression on one side and
then write an equivalent exponential equation.
Example
Solve: log3 x = 2.
log 3 x  2
2
3 x
1
x
2
3
1
x
9
1
Check: x 
9
log 3 x  2
1
log 3
? 2
9
log 3 32
2
 2 TRUE
1
.
The solution is
9
Example
Solve: log x  log x  3  1.
log x  log x  3  1
log  x x  3  1
x x  3  101
x 2  3x  10
x 2  3x  10  0
x  2 x  5   0
x  2  0 or x  5  0
x  2 or x  5
Example (continued)
Check x = 2:
log x  log x  3  1
log 2  log 2  3 ? 1
Check x = –5:
log x  log x  3  1
log 5   log 5  3 ? 1
FALSE
log 2  log 5
log 2  5 
log10
1
1 TRUE
The number –5 is not a solution because negative
numbers do not have real number logarithms. The
solution is 2.
Example
Solve: ln 4x  6   ln x  3  ln x.
ln 4x  6   ln x  3  ln x
4x  6
ln
 ln x
x5
4x  6
x
x5
4x  6
 x x  5 
x  5 
x5
4x  6  x 2  5x
0  x2  x  6
0  x  3x  2 
x  3  0 or x  2  0
x  3 or x  2
Only the value 2 checks
and it is the only solution.
Example - Using the Graphing
Calculator
Solve: e0.5x – 7.3 = 2.08x + 6.2.
Graph y1 = e0.5x – 7.3
and y2 = 2.08x + 6.2
and use the Intersect
method.
The approximate
solutions are –6.471
and 6.610.