Solve for x 5 log base 2 of x=- log base 2 of 32

$5 \log_{2} (x) = - \log_{2} (32)$

Simplify.

Simplify $5 \log_{2} (x)$ by moving $5$ inside the logarithm.

$\log_{2} (x^{5}) = - \log_{2} (32)$

Simplify $- \log_{2} (32)$ by moving $- 1$ inside the logarithm.

$\log_{2} (x^{5}) = \log_{2} (32^{- 1})$

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{5} = 32^{- 1}$

Solve for $x$ .

Take the 5th root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[5]{32^{- 1}}$

Simplify $\sqrt[5]{32^{- 1}}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \sqrt[5]{\frac{1}{32}}$

Rewrite $\sqrt[5]{\frac{1}{32}}$ as $\frac{\sqrt[5]{1}}{\sqrt[5]{32}}$ .

$x = \frac{\sqrt[5]{1}}{\sqrt[5]{32}}$

Any root of $1$ is $1$ .

$x = \frac{1}{\sqrt[5]{32}}$

Simplify the denominator.

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

$x = \frac{1}{\sqrt[5]{2^{5}}}$

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

$x = \frac{1}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{2}$

Decimal Form:

$x = 0.5$