Precalculus Examples

$\frac{1}{2} \cdot (\log_{6} (x) + \log_{6} (y)) - 5 \log_{6} (x + 8)$

Step 1

Simplify each term.

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Step 1.1

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

$\frac{1}{2} \cdot \log_{6} (x y) - 5 \log_{6} (x + 8)$

Step 1.2

Simplify $\frac{1}{2} \log_{6} (x y)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{6} ({(x y)}^{\frac{1}{2}}) - 5 \log_{6} (x + 8)$

Step 1.3

Apply the product rule to $x y$ .

$\log_{6} (x^{\frac{1}{2}} y^{\frac{1}{2}}) - 5 \log_{6} (x + 8)$

Step 1.4

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

$\log_{6} (x^{\frac{1}{2}} y^{\frac{1}{2}}) - \log_{6} ({(x + 8)}^{5})$

Step 2

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

$\log_{6} (\frac{x^{\frac{1}{2}} y^{\frac{1}{2}}}{{(x + 8)}^{5}})$

Step 3

Rewrite $\log_{6} (\frac{x^{\frac{1}{2}} y^{\frac{1}{2}}}{{(x + 8)}^{5}})$ using the change of base formula.

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Step 3.1

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 $0$ .

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

Step 3.2

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

$\frac{\log (\frac{x^{\frac{1}{2}} y^{\frac{1}{2}}}{{(x + 8)}^{5}})}{\log (6)}$