## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x^2\sqrt{x}$
Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ then \begin{array}{l}\require{cancel} \dfrac{\sqrt{x^5}}{\sqrt{x}} \\\\= \dfrac{x^{5/2}}{x^{1/3}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} \dfrac{x^{5/2}}{x^{1/3}} \\\\= x^{\frac{5}{2}-\frac{1}{3}} \\\\= x^{\frac{15}{6}-\frac{2}{6}} \\\\= x^{\frac{13}{6}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ then \begin{array}{l}\require{cancel} x^{\frac{13}{6}} \\\\= \sqrt{x^{13}} .\end{array} Extracting the factor that is a perfect power of the index, then \begin{array}{l}\require{cancel} \sqrt{x^{13}} \\\\= \sqrt{x^{12}\cdot x} \\\\= \sqrt{(x^{2})^6\cdot x} \\\\= x^2\sqrt{x} .\end{array}