Precalculus Examples

$2 \log_{6} (3) + \frac{1}{4} \cdot \log_{6} (r - 8) - \frac{1}{2} \cdot \log_{6} (r)$

Step 1

Simplify each term.

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Step 1.1

Simplify $2 \log_{6} (3)$ by moving $2$ inside the logarithm.

$\log_{6} (3^{2}) + \frac{1}{4} \cdot \log_{6} (r - 8) - \frac{1}{2} \cdot \log_{6} (r)$

Step 1.2

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

$\log_{6} (9) + \frac{1}{4} \cdot \log_{6} (r - 8) - \frac{1}{2} \cdot \log_{6} (r)$

Step 1.3

Simplify $\frac{1}{4} \log_{6} (r - 8)$ by moving $\frac{1}{4}$ inside the logarithm.

$\log_{6} (9) + \log_{6} ({(r - 8)}^{\frac{1}{4}}) - \frac{1}{2} \cdot \log_{6} (r)$

Step 1.4

Multiply $- \frac{1}{2} \log_{6} (r)$ .

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Step 1.4.1

Reorder $\log_{6} (r)$ and $\frac{1}{2}$ .

$\log_{6} (9) + \log_{6} ({(r - 8)}^{\frac{1}{4}}) - (\frac{1}{2} \log_{6} (r))$

Step 1.4.2

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

$\log_{6} (9) + \log_{6} ({(r - 8)}^{\frac{1}{4}}) - \log_{6} (r^{\frac{1}{2}})$

Step 2

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

$\log_{6} (9 {(r - 8)}^{\frac{1}{4}}) - \log_{6} (r^{\frac{1}{2}})$

Step 3

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

$\log_{6} (\frac{9 {(r - 8)}^{\frac{1}{4}}}{r^{\frac{1}{2}}})$