## Intermediate Algebra for College Students (7th Edition)

$\sqrt[20]{x}$
RECALL: (1) $\sqrt[n]{a} = a^{\frac{1}{n}}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ Use rule (1) above to obtain: $=\dfrac{x^{\frac{1}{4}}}{x^{\frac{1}{5}}}$ Use rule (2) above to obtain: $=x^{\frac{1}{4}-\frac{1}{5}}$ Make the fractions similar using their LCD of $20$ to obtain: $=x^{\frac{5}{20} - \frac{4}{20}} \\=x^{\frac{5-4}{20}} \\=x^{\frac{1}{20}}$ Use rule (1) above to obtain: $=\sqrt[20]{x}$