Section 5.5 Solving Exponential and Logarithmic Equations Copyright ©2013, 2009, 2006, 2001 Pearson...
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Transcript of Section 5.5 Solving Exponential and Logarithmic Equations Copyright ©2013, 2009, 2006, 2001 Pearson...
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 PropertyFor any a > 0, a 1,
ax = ay x = y.
Example
Solve:
Write each side as a power of the same number (base).
23x 7 32.
Since the bases are the same number, 2, we can use the base-exponent property and set the exponents equal:
23x 7 25
3x 7 53x 12x 4
Check x = 4:
TRUE32 32
23 4 7 ? 3223x 7 32.
212 7
25
The solution is 4.
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.
This is an exact answer. We cannot simplify further, but we can approximate using a calculator.
log 3x log20
x log 3 log20
x log20log 3
3x 20
2.7268
We can check by finding 32.7268 20.
Example
Solve: 100e0.08t = 2500.
The solution is about 40.2.
0.08100 2500te
0.08ln ln 25te0.08 ln 25t
ln 250.08
t
40.2t
Example
Solve: 4x+3 = 3-x.34 3 x x
3log4 log3 x x
( 3) log4 log3 x xlog4 log3 3log4 x x(log4 log3) 3log4 x
3log4log4 log3
x
1.6737x
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.
log3 x 2
3 2 x
132 x
19
x
The solution is19
.
TRUE 2 2
log319
? 2
log3 3 2
log3 x 2
x 19
Check:
Example
Solve: log x log x 3 1.log x log x 3 1
log x x 3 1
x x 3 101
x2 3x 10x2 3x 10 0x 2 x 5 0
x 2 0 or x 5 0x 2 or x 5
Example (continued)
Check x = –5:
FALSE
log x log x 3 1log 5 log 5 3 ? 1
Check x = 2:
TRUE1 1
log x log x 3 1
log2 log 2 3 ? 1log2 log5
log 25 log10
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
ln4x 6x 5
ln x
4x 6x 5
x
x 5 4x 6x 5
x x 5
4x 6 x2 5x
0 x2 x 60 x 3 x 2
x 3 0 or x 2 0x 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 + 6.2and use the Intersect method.
The approximate solutions are –6.471 and 6.610.