Precalculus Examples

$\frac{\log_{2} (16 - \log (p))}{\log_{2} ({(\sqrt{2})}^{8 - \log (0.0001)})}$

Step 1

Log base $10$ of $0.0001$ is approximately $- 4$ .

$\frac{\log_{2} (16 - \log (p))}{\log_{2} ({\sqrt{2}}^{8 - - 4})}$

Step 2

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

$\frac{\log_{2} (16 - \log (p))}{\log_{2} ({(2^{\frac{1}{2}})}^{8 - - 4})}$

Step 3

Expand $\log_{2} ({(2^{\frac{1}{2}})}^{8 - - 4})$ by moving $8 - - 4$ outside the logarithm.

$\frac{\log_{2} (16 - \log (p))}{(8 - - 4) \log_{2} (2^{\frac{1}{2}})}$

Step 4

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

$\frac{\log_{2} (16 - \log (p))}{(8 - - 4) (\frac{1}{2} \log_{2} (2))}$

Step 5

Logarithm base $2$ of $2$ is $1$ .

$\frac{\log_{2} (16 - \log (p))}{(8 - - 4) (\frac{1}{2} \cdot 1)}$

Step 6

Multiply $- 1$ by $- 4$ .

$\frac{\log_{2} (16 - \log (p))}{(8 + 4) (\frac{1}{2} \cdot 1)}$

Step 7

Multiply $\frac{1}{2}$ by $1$ .

$\frac{\log_{2} (16 - \log (p))}{(8 + 4) \frac{1}{2}}$

Step 8

Add $8$ and $4$ .

$\frac{\log_{2} (16 - \log (p))}{12 (\frac{1}{2})}$

Step 9

Cancel the common factor of $2$ .

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Step 9.1

Factor $2$ out of $12$ .

$\frac{\log_{2} (16 - \log (p))}{2 (6) \frac{1}{2}}$

Step 9.2

Cancel the common factor.

$\frac{\log_{2} (16 - \log (p))}{2 \cdot 6 \frac{1}{2}}$

Step 9.3

Rewrite the expression.

$\frac{\log_{2} (16 - \log (p))}{6}$