Precalculus Examples

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

Step 1

Simplify each term.

Tap for more steps...

Step 1.1

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

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

Step 1.2

Multiply $4$ by $16$ .

$\frac{1}{2} \cdot \log_{2} (64) - \log_{2} (2)$

Step 1.3

Logarithm base $2$ of $64$ is $6$ .

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

Step 1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.4.1

Factor $2$ out of $6$ .

$\frac{1}{2} \cdot (2 (3)) - \log_{2} (2)$

Step 1.4.2

Cancel the common factor.

$\frac{1}{2} \cdot (2 \cdot 3) - \log_{2} (2)$

Step 1.4.3

Rewrite the expression.

$3 - \log_{2} (2)$

Step 1.5

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

$3 - 1 \cdot 1$

Step 1.6

Multiply $- 1$ by $1$ .

$3 - 1$

Step 2

Subtract $1$ from $3$ .

$2$