Precalculus Examples

$\frac{\log_{2} (200)}{\log_{2} (2 b)}$

Step 1

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

$\frac{\log_{2} (200)}{\log_{2} (2) + \log_{2} (b)}$

Step 2

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

$\frac{\log_{2} (200)}{1 + \log_{2} (b)}$

Step 3

Simplify the numerator.

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Step 3.1

Rewrite $\log_{2} (200)$ as $\log_{2} (2^{3} \cdot 5^{2})$ .

$\frac{\log_{2} (2^{3} \cdot 5^{2})}{1 + \log_{2} (b)}$

Step 3.2

Rewrite $\log_{2} (2^{3} \cdot 5^{2})$ as $\log_{2} (2^{3}) + \log_{2} (5^{2})$ .

$\frac{\log_{2} (2^{3}) + \log_{2} (5^{2})}{1 + \log_{2} (b)}$

Step 3.3

Use logarithm rules to move $3$ out of the exponent.

$\frac{3 \log_{2} (2) + \log_{2} (5^{2})}{1 + \log_{2} (b)}$

Step 3.4

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

$\frac{3 \cdot 1 + \log_{2} (5^{2})}{1 + \log_{2} (b)}$

Step 3.5

Multiply $3$ by $1$ .

$\frac{3 + \log_{2} (5^{2})}{1 + \log_{2} (b)}$

Step 3.6

Expand $\log_{2} (5^{2})$ by moving $2$ outside the logarithm.

$\frac{3 + 2 \log_{2} (5)}{1 + \log_{2} (b)}$