## Precalculus (6th Edition) Blitzer

$x=16$
RECALL: (i) $\log_b{x}=y \longrightarrow b^y=x.$ (ii) $a^{\frac{m}{n}}=(\sqrt[n]{a})^m.$ Use rule (i) above to write the given equation in exponential form, then solve the equation: $\begin{array}{ccc} &64^{\frac{2}{3}} &= &x\end{array}$ use rule (ii) above to obtain $\begin{array}{ccc} \\(\sqrt[3]{64})^2 &= &x \\(\sqrt[3]{4^3})^2 &= &x \\(4)^2 &= &x \\16 &= &x \end{array}.$ Thus, $x=16$.