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# Exponents and Logarithmic Equations Help

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Introduction to Exponents and Logarithmic Equations

For some logarithmic equations, a solution might be extraneous solution. That is, such a solution is a solution to the rewritten equations but not to the original equations. Some solutions to the rewritten equations will cause logarithms of 0 or of negative numbers. We can check them in the original equation to see which solutions are true solutions.

#### Examples

Solve for x .

• log 2 ( x 2 + 3 x − 10) = 3
• We will rewrite this as an exponent equation: 2 3 = x 2 + 3 x − 10 and solve for x .

x 2 + 3 x − 10 = 8

x 2 + 3 x − 18 = 0

( x + 6)( x − 3) = 0

The solutions are x = − 6 and x = 3. We will check them in the original equation.

The solutions to the original equation are x = −6 and x = 3.

• log 5 ( x 2 + 5 x − 4) = log 5 ( x + 1)
• The logs cancel leaving x 2 + 5 x − 4 = x + 1.

x 2 + 5 x − 4 = x + 1

x 2 + 4 x − 5 = 0

( x + 5)( x − 1) = 0

The solutions are x = − 5 and x = 1. We cannot allow x = − 5 as a solution because log 5 (−5 + 1) is not defined. We need to check x = 1.

log 5 (1 2 + 5(1) − 4) = log 5 (1 + 1) is true

The solution is x = 1.

### Exponents and Logarithmic Equations Practice Problems

#### Practice

Solve for x .

1. ln( x 2 + x − 20) = ln(3 x + 4)
2. log 4 (2 x 2 − 3 x + 59) = 3

#### Solutions

1. ln( x 2 + x − 20) = ln(3 x + 4)

x 2 + x − 20 = 3 x + 4

x 2 − 2 x − 24 = 0

( x − 6)( x + 4) = 0

The solutions are x = 6 and x = −4. Because ln[3(−4) + 4] is not defined, we only need to check x = 6.

ln(6 2 + 6 − 20) = ln[3(6) + 4] is true.

The only solution is x = 6.

2. log 4 (2 x 2 − 3 x + 59) = 3

## Three More Important Logarithm Properties

The following three logarithm properties come directly from the exponent properties .

1. log b mn = log b m + log b n
2. log b m t = t log b m

We will see why Property 1 works. Let x = log b m and y = log b n . Rewriting these equations as exponential equations, we get b x = m and b y = n . Multiplying m and n , we have mn = b x · b y = b x + y . Rewriting the equation mn = b x + y as a logarithmic equation, we get log b mn = x + y . Because x = log b m and y = log b n , log b mn = x + y becomes log b mn = log b m + log b n .

#### Examples

Use Property 1 to rewrite the logarithms.

• log 4 7 x = log 4 7 + log 4 x
• ln 15 t = ln 15 + ln t
• log 6 19 t 2 = log 6 19 + log 6 t 2
• log 100 y 4 = log 10 2 + log y 4 = 2 + log y 4
• log 9 3 + log 9 27 = log 9 3(27) = log 9 81 = 2

Use Property 2 to rewrite the logarithms.

The exponent property allows us to apply the third logarithm property to roots as well as to powers. The third logarithm property is especially useful in science and business applications.

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