Algebra Examples

$4 \log_{2} (16) + 5 \log_{2} (\frac{1}{32}) + \log_{5} (\sqrt{5})$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $2$ of $16$ is $4$ .

$4 \cdot 4 + 5 \log_{2} (\frac{1}{32}) + \log_{5} (\sqrt{5})$

Step 1.2

Multiply $4$ by $4$ .

$16 + 5 \log_{2} (\frac{1}{32}) + \log_{5} (\sqrt{5})$

Step 1.3

Logarithm base $2$ of $\frac{1}{32}$ is $- 5$ .

$16 + 5 \cdot - 5 + \log_{5} (\sqrt{5})$

Step 1.4

Multiply $5$ by $- 5$ .

$16 - 25 + \log_{5} (\sqrt{5})$

Step 2

Subtract $25$ from $16$ .

$- 9 + \log_{5} (\sqrt{5})$

Step 3

Rewrite $\log_{5} (\sqrt{5})$ using the change of base formula.

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Step 3.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$- 9 + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 3.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$- 9 + \frac{\log (\sqrt{5})}{\log (5)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- 9 + \frac{\log (\sqrt{5})}{\log (5)}$

Decimal Form:

$- 8.5$