Algebra 1: Common Core (15th Edition)

The answer is $a^{\frac{1}{6}}$
To solve this $a^{\frac{2}{3}}$$\times$$a^{x}$ = $a^{\frac{5}{6}}$ $a^{x}$ = $\frac{a^{\frac{5}{6}}}{a^{\frac{2}{3}}}$ $a^{x}$ = $a^{\frac{5}{6}-\frac{2}{3}}$ $a^{x}$ = $a^{\frac{5}{6}-\frac{4}{6}}$ $a^{x}$ = $a^{\frac{1}{6}}$ --> since bases are equal, make the exponents equal x = $\frac{1}{6}$ Therefore the missing part is $a^{\frac{1}{6}}$.