Precalculus Examples

$\log_{8} (\frac{4 x + 2}{\sqrt[6]{2 - 3 x}})$

Step 1

Rewrite $\log_{8} (\frac{4 x + 2}{\sqrt[6]{2 - 3 x}})$ as $\log_{8} (4 x + 2) - \log_{8} (\sqrt[6]{2 - 3 x})$ .

$\log_{8} (4 x + 2) - \log_{8} (\sqrt[6]{2 - 3 x})$

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[6]{2 - 3 x}$ as ${(2 - 3 x)}^{\frac{1}{6}}$ .

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

Step 3

Expand $\log_{8} ({(2 - 3 x)}^{\frac{1}{6}})$ by moving $\frac{1}{6}$ outside the logarithm.

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

Step 4

Combine $\frac{1}{6}$ and $\log_{8} (2 - 3 x)$ .

$\log_{8} (4 x + 2) - \frac{\log_{8} (2 - 3 x)}{6}$

Step 5

To write $\log_{8} (4 x + 2)$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

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

Step 6

Combine $\log_{8} (4 x + 2)$ and $\frac{6}{6}$ .

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

Step 7

Combine the numerators over the common denominator.

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

Step 8

Move $6$ to the left of $\log_{8} (4 x + 2)$ .

$\frac{6 \log_{8} (4 x + 2) - \log_{8} (2 - 3 x)}{6}$