Precalculus Examples

$2 \log_{8} (\sqrt[5]{64})$

Step 1

Rewrite $64$ as $2^{5} \cdot 2$ .

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Step 1.1

Factor $32$ out of $64$ .

$2 \log_{8} (\sqrt[5]{32 (2)})$

Step 1.2

Rewrite $32$ as $2^{5}$ .

$2 \log_{8} (\sqrt[5]{2^{5} \cdot 2})$

Step 2

Pull terms out from under the radical.

$2 \log_{8} (2 \sqrt[5]{2})$

Step 3

Simplify $2 \log_{8} (2 \sqrt[5]{2})$ by moving $2$ inside the logarithm.

$\log_{8} ({(2 \sqrt[5]{2})}^{2})$

Step 4

Apply the product rule to $2 \sqrt[5]{2}$ .

$\log_{8} (2^{2} {\sqrt[5]{2}}^{2})$

Step 5

Raise $2$ to the power of $2$ .

$\log_{8} (4 {\sqrt[5]{2}}^{2})$

Step 6

Rewrite ${\sqrt[5]{2}}^{2}$ as $\sqrt[5]{2^{2}}$ .

$\log_{8} (4 \sqrt[5]{2^{2}})$

Step 7

Raise $2$ to the power of $2$ .

$\log_{8} (4 \sqrt[5]{4})$

Step 8

Rewrite $\log_{8} (4 \sqrt[5]{4})$ using the change of base formula.

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Step 8.1

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 $0$ .

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

Step 8.2

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

$\frac{\log (4 \sqrt[5]{4})}{\log (8)}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (4 \sqrt[5]{4})}{\log (8)}$

Decimal Form:

$0.8$