Precalculus Examples

$\frac{1}{4} \cdot (\log_{8} (x) + \log_{8} (y)) - 5 \log_{8} (x + 6)$

Step 1

Simplify each term.

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Step 1.1

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

$\frac{1}{4} \cdot \log_{8} (x y) - 5 \log_{8} (x + 6)$

Step 1.2

Simplify $\frac{1}{4} \log_{8} (x y)$ by moving $\frac{1}{4}$ inside the logarithm.

$\log_{8} ({(x y)}^{\frac{1}{4}}) - 5 \log_{8} (x + 6)$

Step 1.3

Apply the product rule to $x y$ .

$\log_{8} (x^{\frac{1}{4}} y^{\frac{1}{4}}) - 5 \log_{8} (x + 6)$

Step 1.4

Simplify $- 5 \log_{8} (x + 6)$ by moving $5$ inside the logarithm.

$\log_{8} (x^{\frac{1}{4}} y^{\frac{1}{4}}) - \log_{8} ({(x + 6)}^{5})$

Step 2

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

$\log_{8} (\frac{x^{\frac{1}{4}} y^{\frac{1}{4}}}{{(x + 6)}^{5}})$