Chapter 11 - Additional Topics - 11.3 - Fractional Exponents - Problem Set 11.3 - Page 489: 59

$2y^{\frac{5}{12}}$

We will use the rule $a^{n}a^{m}=a^{n+m}$ in the solution. Therefore, $(2y^{\frac{2}{3}})(y^{-\frac{1}{4}})$ $=(2\times y^{\frac{2}{3}})(1\times y^{-\frac{1}{4}})$ $=(2\times1)\times(y^{\frac{2}{3}}\times y^{-\frac{1}{4}})$ $=(2)(y^{\frac{2}{3}-\frac{1}{4}})$ $=(2)(y^{\frac{2(4)-3}{12}})$ $=2y^{\frac{5}{12}}$