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: 95

Answer

$y^{5/6}z^{1/3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents and the laws of exponents to simplify the given expression, $ \sqrt{y}\cdot\sqrt[3]{yz} .$ $\bf{\text{Solution Details:}}$ 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} y^{1/2}\cdot(yz)^{1/3} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} y^{1/2}\cdot(y^{1/3}z^{1/3}) \\\\= y^{1/2}y^{1/3}z^{1/3} .\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} y^{\frac{1}{2}+\frac{1}{3}}z^{\frac{1}{3}} .\end{array} To simplify the expression $ \dfrac{1}{2}+\dfrac{1}{3} ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $ 2 $ and $ 3 $ is $ 6 $ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} y^{\frac{1}{2}\cdot\frac{3}{3}+\frac{1}{3}\cdot\frac{2}{2}}z^{\frac{1}{3}} \\\\= y^{\frac{3}{6}+\frac{2}{6}}z^{\frac{1}{3}} \\\\= y^{\frac{5}{6}}z^{\frac{1}{3}} \\\\= y^{5/6}z^{1/3} .\end{array}
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