Precalculus Examples

$\log_{9} (\frac{9 \sqrt{7}}{2 n})$

Step 1

Rewrite $\log_{9} (\frac{9 \sqrt{7}}{2 n})$ as $\log_{9} (9 \sqrt{7}) - \log_{9} (2 n)$ .

$\log_{9} (9 \sqrt{7}) - \log_{9} (2 n)$

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{7}$ as $7^{\frac{1}{2}}$ .

$\log_{9} (9 \cdot 7^{\frac{1}{2}}) - \log_{9} (2 n)$

Step 3

Rewrite $\log_{9} (9 \cdot 7^{\frac{1}{2}})$ as $\log_{9} (9) + \log_{9} (7^{\frac{1}{2}})$ .

$\log_{9} (9) + \log_{9} (7^{\frac{1}{2}}) - \log_{9} (2 n)$

Step 4

Expand $\log_{9} (7^{\frac{1}{2}})$ by moving $\frac{1}{2}$ outside the logarithm.

$\log_{9} (9) + \frac{1}{2} \log_{9} (7) - \log_{9} (2 n)$

Step 5

Logarithm base $9$ of $9$ is $1$ .

$1 + \frac{1}{2} \log_{9} (7) - \log_{9} (2 n)$

Step 6

Combine $\frac{1}{2}$ and $\log_{9} (7)$ .

$1 + \frac{\log_{9} (7)}{2} - \log_{9} (2 n)$

Step 7

Rewrite $\log_{9} (2 n)$ as $\log_{9} (2) + \log_{9} (n)$ .

$1 + \frac{\log_{9} (7)}{2} - (\log_{9} (2) + \log_{9} (n))$

Step 8

Apply the distributive property.

$1 + \frac{\log_{9} (7)}{2} - \log_{9} (2) - \log_{9} (n)$