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 - Page 449: 70



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $ \dfrac{z^{3/4}}{z^{5/4}\cdot z^{-2}} .$ $\bf{\text{Solution Details:}}$ 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} \dfrac{z^{3/4}}{z^{\frac{5}{4}+(-2)}} \\\\= \dfrac{z^{3/4}}{z^{\frac{5}{4}-2}} .\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} z^{\frac{3}{4}-\left( \frac{5}{4}-2 \right)} \\\\= z^{\frac{3}{4}-\frac{5}{4}+2} \\\\= z^{-\frac{2}{4}+2} \\\\= z^{-\frac{1}{2}+2} .\end{array} To simplify the expression $ -\dfrac{1}{2}+2 ,$ change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $ 2 $ and $ 1 $ is $ 2 $ 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} z^{-\frac{1}{2}+2\cdot\frac{2}{2}} \\\\= z^{-\frac{1}{2}+\frac{4}{2}} \\\\= z^{\frac{3}{2}} \\\\= z^{3/2} .\end{array}
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