## Intermediate Algebra (12th Edition)

F. $\sqrt 6$
As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $6^{\frac{2}{4}}=6^{\frac{2\div2}{4\div2}}=6^{\frac{1}{2}}=\sqrt[2] 6^{1}=\sqrt 6$.