Precalculus Examples

$\log_{2} (16) + \log_{2} (64)$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} (16 \cdot 64)$

Step 2

Multiply $16$ by $64$ .

$\log_{2} (1024)$

Step 3

Logarithm base $2$ of $1024$ is $10$ .

$10$