Precalculus Examples

$\log_{0.25} (\sqrt[3]{16})$

Step 1

Rewrite $16$ as $2^{3} \cdot 2$ .

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Step 1.1

Factor $8$ out of $16$ .

$\log_{0.25} (\sqrt[3]{8 (2)})$

Step 1.2

Rewrite $8$ as $2^{3}$ .

$\log_{0.25} (\sqrt[3]{2^{3} \cdot 2})$

Step 2

Pull terms out from under the radical.

$\log_{0.25} (2 \sqrt[3]{2})$

Step 3

Rewrite $\log_{0.25} (2 \sqrt[3]{2})$ 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 $0$ .

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

Step 3.2

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

$\frac{\log (2 \sqrt[3]{2})}{\log (0.25)}$

Step 4

Log base $10$ of $0.25$ is approximately $- 0.60205999$ .

$\frac{\log (2 \sqrt[3]{2})}{- 0.60205999}$

Step 5

Move the negative in front of the fraction.

$- \frac{\log (2 \sqrt[3]{2})}{0.60205999}$

Step 6

Evaluate the root.

$- \frac{\log (2 \cdot 1.25992104)}{0.60205999}$

Step 7

Multiply $2$ by $1.25992104$ .

$- \frac{\log (2.51984209)}{0.60205999}$

Step 8

Log base $10$ of $2.51984209$ is approximately $0.40137332$ .

$- \frac{0.40137332}{0.60205999}$

Step 9

Divide $0.40137332$ by $0.60205999$ .

$- 1 \cdot 0.66666666$

Step 10

Multiply $- 1$ by $0.66666666$ .

$- 0.66666666$