## Precalculus (6th Edition) Blitzer

$\log_{64}{4}=\frac{1}{3}$
RECALL: (i) $\sqrt[n]{a}=a^{\frac{1}{n}}.$ (ii) $b^y=x \longleftrightarrow \log_b{x} =y .$ Use rule (i) above to obtain $64^{\frac{1}{3}}=4.$ Use rule (ii) above to obtain $64^{\frac{1}{3}} = 4 \longrightarrow \log_{64}{4}=\frac{1}{3}.$ Thus, the equivalent logarithmic form of the given expression is $\log_{64}{4}=\frac{1}{3}.$