Evaluate log base 8 of cube root of 1/64

$\log_{8} (\sqrt[3]{\frac{1}{64}})$

Rewrite $\sqrt[3]{\frac{1}{64}}$ as $\frac{\sqrt[3]{1}}{\sqrt[3]{64}}$ .

$\log_{8} (\frac{\sqrt[3]{1}}{\sqrt[3]{64}})$

Any root of $1$ is $1$ .

$\log_{8} (\frac{1}{\sqrt[3]{64}})$

Simplify the denominator.

Rewrite $64$ as $4^{3}$ .

$\log_{8} (\frac{1}{\sqrt[3]{4^{3}}})$

Pull terms out from under the radical, assuming real numbers.

$\log_{8} (\frac{1}{4})$

Logarithm base $8$ of $\frac{1}{4}$ is $- \frac{2}{3}$ .

Rewrite as an equation.

$\log_{8} (\frac{1}{4}) = x$

Rewrite $\log_{8} (\frac{1}{4}) = x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ does not equal $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$8^{x} = \frac{1}{4}$

Create expressions in the equation that all have equal bases.

${(2^{3})}^{x} = 2^{- 2}$

Rewrite ${(2^{3})}^{x}$ as $2^{3 x}$ .

$2^{3 x} = 2^{- 2}$

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$3 x = - 2$

Solve for $x$ .

$x = - \frac{2}{3}$

The variable $x$ is equal to $- \frac{2}{3}$ .

$- \frac{2}{3}$

The result can be shown in multiple forms.

Exact Form:

$- \frac{2}{3}$

Decimal Form:

$- 0.\overline{6}$

Evaluate log base 8 of cube root of 1/64

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