## Intermediate Algebra (12th Edition)

$\frac{8}{9}$
As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $(\frac{64}{81})^{\frac{1}{2}}=\sqrt[2] (\frac{64}{81})^{1}=\sqrt (\frac{64}{81})=\frac{8}{9}$. We know that $\sqrt (\frac{64}{81})=\frac{8}{9}$, because $(\frac{8}{9})^{2}=(\frac{64}{81})$.