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# Simple Exponent and Logarithm Equations Help (page 3)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

#### Examples

• f ( x ) = log 5 (2 − x )
• Because we are taking the log of 2 − x , 2 − x needs to be positive.

2 − x > 0

x > − 2

x < 2

The domain is (−∞, 2).

• f ( x ) = log( x 2x − 2)

x 2x − 2 > 0

( x − 2)( x + 1) > 0

Put x = 2 and x = − 1 on the number line and test to see where ( x − 2) ( x + 1) > 0 is true.

Fig. 9.13

We want the “True” intervals, so the domain is (−∞, −1) ∪ (2, ∞).

• g ( x ) = ln( x 2 + 1)

Because x 2 + 1 is always positive, the domain is all real numbers, (−∞, ∞).

### Finding the Domain of a Function Practice Problems

#### Practice

1. f ( x ) = ln(10 − 2 x )
2. h ( x ) = log( x 2 − 4)
3. f ( x ) = log( x 2 + 4)

#### Solutions

1. Solve 10 − 2 x > 0. The domain is x < 5, (−∞, 5).
2. Solve x 2 − 4 > 0

Fig. 9.14

The domain is (−∞, −2) ∪ (2, ∞).

1. Because x 2 + 4 > 0 is always positive, the domain is all real numbers, (—∞, ∞).

Find practice problems and solutions for these concepts at: Exponents and Logarithms Practice Test.

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