Precalculus Examples

$\frac{1}{2} \cdot \log_{8} (w) + 7 \log_{8} (z) - \log_{8} (y)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{2} \log_{8} (w)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{8} (w^{\frac{1}{2}}) + 7 \log_{8} (z) - \log_{8} (y)$

Step 1.2

Simplify $7 \log_{8} (z)$ by moving $7$ inside the logarithm.

$\log_{8} (w^{\frac{1}{2}}) + \log_{8} (z^{7}) - \log_{8} (y)$

Step 2

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

$\log_{8} (w^{\frac{1}{2}} z^{7}) - \log_{8} (y)$

Step 3

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

$\log_{8} (\frac{w^{\frac{1}{2}} z^{7}}{y})$

Step 4

Reorder factors in $\log_{8} (\frac{w^{\frac{1}{2}} z^{7}}{y})$ .

$\log_{8} (\frac{z^{7} w^{\frac{1}{2}}}{y})$