Algebra Examples

\frac{1}{2} ({log}_{2} (4) + {log}_{2} (16)) - {log}_{2} (2)

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

Step 1

Simplify each term.

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Step 1.1

Use the product property of logarithms,

{log}_{b} (x) + {log}_{b} (y) = {log}_{b} (x y)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

\frac{1}{2} \cdot {log}_{2} (4 \cdot 16) - {log}_{2} (2)

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

Step 1.2

Multiply

4

$4$ by

16

$16$ .

\frac{1}{2} \cdot {log}_{2} (64) - {log}_{2} (2)

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

Step 1.3

Logarithm base

2

$2$ of

64

$64$ is

6

$6$ .

\frac{1}{2} \cdot 6 - {log}_{2} (2)

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

Step 1.4

Cancel the common factor of

2

$2$ .

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Step 1.4.1

Factor

2

$2$ out of

6

$6$ .

\frac{1}{2} \cdot (2 (3)) - {log}_{2} (2)

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

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

$3$ .

$2$