Precalculus Examples

$\frac{1}{2} \cdot (\ln (x + 2) - \ln (x^{2} - 4))$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{1}{2} \cdot \ln (\frac{x + 2}{x^{2} - 4})$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $4$ as $2^{2}$ .

$\frac{1}{2} \cdot \ln (\frac{x + 2}{x^{2} - 2^{2}})$

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$\frac{1}{2} \cdot \ln (\frac{x + 2}{(x + 2) (x - 2)})$

Step 3

Cancel the common factor of $x + 2$ .

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Step 3.1

Cancel the common factor.

$\frac{1}{2} \cdot \ln (\frac{x + 2}{(x + 2) (x - 2)})$

Step 3.2

Rewrite the expression.

$\frac{1}{2} \cdot \ln (\frac{1}{x - 2})$

Step 4

Simplify $\frac{1}{2} \ln (\frac{1}{x - 2})$ by moving $\frac{1}{2}$ inside the logarithm.

$\ln ({(\frac{1}{x - 2})}^{\frac{1}{2}})$

Step 5

Apply the product rule to $\frac{1}{x - 2}$ .

$\ln (\frac{1^{\frac{1}{2}}}{{(x - 2)}^{\frac{1}{2}}})$

Step 6

One to any power is one.

$\ln (\frac{1}{{(x - 2)}^{\frac{1}{2}}})$