Solve for x log base x of 243/32=5

$\log_{x} (\frac{243}{32}) = 5$

Rewrite $\log_{x} (\frac{243}{32}) = 5$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$x^{5} = \frac{243}{32}$

Solve for $x$ .

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

$x = \sqrt[5]{\frac{243}{32}}$

Simplify $\sqrt[5]{\frac{243}{32}}$ .

Rewrite $\sqrt[5]{\frac{243}{32}}$ as $\frac{\sqrt[5]{243}}{\sqrt[5]{32}}$ .

$x = \frac{\sqrt[5]{243}}{\sqrt[5]{32}}$

Simplify the numerator.

Rewrite $243$ as $3^{5}$ .

$x = \frac{\sqrt[5]{3^{5}}}{\sqrt[5]{32}}$

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

$x = \frac{3}{\sqrt[5]{32}}$

Simplify the denominator.

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

$x = \frac{3}{\sqrt[5]{2^{5}}}$

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

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.5$

Mixed Number Form:

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

Solve for x log base x of 243/32=5

Related Questions

Support LCMGCF.com by disabling your adblocker.