## Intermediate Algebra (12th Edition)

$y$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $y^{7/3}\cdot y^{-4/3} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} y^{\frac{7}{3}+\left(-\frac{4}{3} \right)} \\\\= y^{\frac{7}{3}-\frac{4}{3}} \\\\= y^{\frac{3}{3}} \\\\= y^{1} \\\\= y .\end{array}