Precalculus Examples

$\log (x) + \log (x^{2} - 49) - \log (2) - \log (x + 7)$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (x (x^{2} - 49)) - \log (2) - \log (x + 7)$

Step 2

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

$\log (\frac{x (x^{2} - 49)}{2}) - \log (x + 7)$

Step 3

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

$\log (\frac{\frac{x (x^{2} - 49)}{2}}{x + 7})$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$\log (\frac{x (x^{2} - 49)}{2} \cdot \frac{1}{x + 7})$

Step 5

Simplify the numerator.

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Step 5.1

Rewrite $49$ as $7^{2}$ .

$\log (\frac{x (x^{2} - 7^{2})}{2} \cdot \frac{1}{x + 7})$

Step 5.2

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 $b = 7$ .

$\log (\frac{x (x + 7) (x - 7)}{2} \cdot \frac{1}{x + 7})$

Step 6

Cancel the common factor of $x + 7$ .

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Step 6.1

Factor $x + 7$ out of $x (x + 7) (x - 7)$ .

$\log (\frac{(x + 7) (x (x - 7))}{2} \cdot \frac{1}{x + 7})$

Step 6.2

Cancel the common factor.

$\log (\frac{(x + 7) (x (x - 7))}{2} \cdot \frac{1}{x + 7})$

Step 6.3

Rewrite the expression.

$\log (\frac{x (x - 7)}{2})$