## Intermediate Algebra (12th Edition)

$k^{\frac{1}{6}}$
$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents and the laws of exponents to simplify the given expression, $\sqrt[3]{\sqrt[]{k}} .$ $\bf{\text{Solution Details:}}$ Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{\sqrt[2]{k}} \\\\= \sqrt[3]{k^{1/2}} \\\\= \left( k^{1/2} \right)^{1/3} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} k^{\frac{1}{2}\cdot\frac{1}{3}} \\\\= k^{\frac{1}{6}} .\end{array}