Precalculus Examples

$4 \log_{a} (6 x^{7}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Step 1

Simplify each term.

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Simplify $4 \log_{a} (6 x^{7})$ by moving $4$ inside the logarithm.

$\log_{a} ({(6 x^{7})}^{4}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Apply the product rule to $6 x^{7}$ .

$\log_{a} (6^{4} {(x^{7})}^{4}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Raise $6$ to the power of $4$ .

$\log_{a} (1296 {(x^{7})}^{4}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Multiply the exponents in ${(x^{7})}^{4}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\log_{a} (1296 x^{7 \cdot 4}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Multiply $7$ by $4$ .

$\log_{a} (1296 x^{28}) - \frac{1}{9} \cdot \log_{a} (2 x + 13)$

Multiply $- \frac{1}{9} \log_{a} (2 x + 13)$ .

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Reorder $\log_{a} (2 x + 13)$ and $\frac{1}{9}$ .

$\log_{a} (1296 x^{28}) - (\frac{1}{9} \log_{a} (2 x + 13))$

Simplify $\frac{1}{9} \log_{a} (2 x + 13)$ by moving $\frac{1}{9}$ inside the logarithm.

$\log_{a} (1296 x^{28}) - \log_{a} ({(2 x + 13)}^{\frac{1}{9}})$

Step 2

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

$\log_{a} (\frac{1296 x^{28}}{{(2 x + 13)}^{\frac{1}{9}}})$