Precalculus Examples

$\log_{3} (x + 7) - \log_{3} (x - 7) + 2$

Step 1

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

$\log_{3} (\frac{x + 7}{x - 7}) + 2$

Step 2

Rewrite $\log_{3} (\frac{x + 7}{x - 7})$ using the change of base formula.

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Step 2.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + 2$

Step 2.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (\frac{x + 7}{x - 7})}{\log (3)} + 2$

Step 3

To write $2$ as a fraction with a common denominator, multiply by $\frac{\log (3)}{\log (3)}$ .

$\frac{\log (\frac{x + 7}{x - 7})}{\log (3)} + \frac{2 \log (3)}{\log (3)}$

Step 4

Combine the numerators over the common denominator.

$\frac{\log (\frac{x + 7}{x - 7}) + 2 \log (3)}{\log (3)}$

Step 5

Simplify the numerator.

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Step 5.1

Simplify $2 \log (3)$ by moving $2$ inside the logarithm.

$\frac{\log (\frac{x + 7}{x - 7}) + \log (3^{2})}{\log (3)}$

Step 5.2

Raise $3$ to the power of $2$ .

$\frac{\log (\frac{x + 7}{x - 7}) + \log (9)}{\log (3)}$

Step 5.3

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

$\frac{\log (\frac{x + 7}{x - 7} \cdot 9)}{\log (3)}$

Step 5.4

Combine $\frac{x + 7}{x - 7}$ and $9$ .

$\frac{\log (\frac{(x + 7) \cdot 9}{x - 7})}{\log (3)}$

Step 5.5

Move $9$ to the left of $x + 7$ .

$\frac{\log (\frac{9 (x + 7)}{x - 7})}{\log (3)}$