Precalculus Examples

$\log_{8} (\frac{7 x + 3}{\sqrt[4]{6 - 3 x}})$

Step 1

Rewrite $\log_{8} (\frac{7 x + 3}{\sqrt[4]{6 - 3 x}})$ as $\log_{8} (7 x + 3) - \log_{8} (\sqrt[4]{6 - 3 x})$ .

$\log_{8} (7 x + 3) - \log_{8} (\sqrt[4]{6 - 3 x})$

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{6 - 3 x}$ as ${(6 - 3 x)}^{\frac{1}{4}}$ .

$\log_{8} (7 x + 3) - \log_{8} ({(6 - 3 x)}^{\frac{1}{4}})$

Step 3

Expand $\log_{8} ({(6 - 3 x)}^{\frac{1}{4}})$ by moving $\frac{1}{4}$ outside the logarithm.

$\log_{8} (7 x + 3) - (\frac{1}{4} \log_{8} (6 - 3 x))$

Step 4

Combine $\frac{1}{4}$ and $\log_{8} (6 - 3 x)$ .

$\log_{8} (7 x + 3) - \frac{\log_{8} (6 - 3 x)}{4}$

Step 5

To write $\log_{8} (7 x + 3)$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\log_{8} (7 x + 3) \cdot \frac{4}{4} - \frac{\log_{8} (6 - 3 x)}{4}$

Step 6

Combine $\log_{8} (7 x + 3)$ and $\frac{4}{4}$ .

$\frac{\log_{8} (7 x + 3) \cdot 4}{4} - \frac{\log_{8} (6 - 3 x)}{4}$

Step 7

Combine the numerators over the common denominator.

$\frac{\log_{8} (7 x + 3) \cdot 4 - \log_{8} (6 - 3 x)}{4}$

Step 8

Move $4$ to the left of $\log_{8} (7 x + 3)$ .

$\frac{4 \log_{8} (7 x + 3) - \log_{8} (6 - 3 x)}{4}$