Precalculus Examples

$6 \log_{7} (8 y + 1) + \frac{1}{4} \cdot \log_{7} (y + 9)$

Step 1

Simplify each term.

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Step 1.1

Simplify $6 \log_{7} (8 y + 1)$ by moving $6$ inside the logarithm.

$\log_{7} ({(8 y + 1)}^{6}) + \frac{1}{4} \cdot \log_{7} (y + 9)$

Step 1.2

Simplify $\frac{1}{4} \log_{7} (y + 9)$ by moving $\frac{1}{4}$ inside the logarithm.

$\log_{7} ({(8 y + 1)}^{6}) + \log_{7} ({(y + 9)}^{\frac{1}{4}})$

Step 2

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

$\log_{7} ({(8 y + 1)}^{6} {(y + 9)}^{\frac{1}{4}})$

Step 3

Reorder factors in $\log_{7} ({(8 y + 1)}^{6} {(y + 9)}^{\frac{1}{4}})$ .

$\log_{7} ({(y + 9)}^{\frac{1}{4}} {(8 y + 1)}^{6})$