Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.2 - Rational Exponents - 7.2 Exercises: 57

Answer

$x^{5/6}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents and the laws of exponents to convert the given expression, $ \sqrt[3]{x}\cdot\sqrt{x} ,$ to exponential form. $\bf{\text{Solution Details:}}$ 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} x^{\frac{1}{3}}\cdot x^{\frac{1}{2}} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^{\frac{1}{3}+\frac{1}{2}} .\end{array} To simplify the expression, $ \dfrac{1}{3}+\dfrac{1}{2} ,$ find the $LCD$ of the denominators $\{ 3,2 \}.$ The $LCD$ is $ 6 $ since it is the lowest number that can be divided by both denominators. Multiplying both the numerator and the denominator of each term by the constant that will make the denominators equal to the $LCD$ results to \begin{array}{l}\require{cancel} x^{\frac{1}{3}\cdot\frac{2}{2}+\frac{1}{2}\cdot\frac{3}{3}} \\\\= x^{\frac{2}{6}+\frac{3}{6}} \\\\= x^{\frac{5}{6}} \\\\= x^{5/6} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.