How do you solve for x in log_x(32)=5?

Aug 21, 2015

If ${\log}_{x} \left(32\right) = 5$ then ${x}^{5} = {x}^{{\log}_{x} \left(32\right)} = 32$.

So $x = \sqrt[5]{32} = \sqrt[5]{{2}^{5}} = 2$.

Explanation:

Alternatively, use the change of base formula:

$5 = {\log}_{x} \left(32\right) = \ln \frac{32}{\ln} \left(x\right)$

So $\ln \left(x\right) = \ln \frac{32}{5} = \ln \left(\sqrt[5]{32}\right) = \ln \left(\sqrt[5]{{2}^{5}}\right) = \ln \left(2\right)$

Hence $x = 2$