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.