## Intermediate Algebra (12th Edition)

$s^{\frac{1}{2}}$
We are given that $(\sqrt[n] a)^{m}=\sqrt[n] a^{m}=a^{\frac{m}{n}}$, as long as all indicated roots are real numbers. Therefore, $\sqrt[8] s^{4}=s^{\frac{4}{8}}=s^{\frac{4\div4}{8\div4}}=s^{\frac{1}{2}}$.