Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 5 - Number Theory and the Real Number System - 5.6 Exponents and Scientific Notation - Exercise Set 5.6 - Page 320: 42

Answer

\[\frac{-3{{y}^{15}}}{{{x}^{3}}}\].

Work Step by Step

Quotient rule: When exponential expressions with same base are divided then subtract the exponent in the denominator with the exponent in the numerator of the common base: \[\frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\]. So, \[\begin{align} & \frac{24{{x}^{2}}{{y}^{13}}}{-8{{x}^{5}}{{y}^{-2}}}=-\frac{24}{8}\left( {{x}^{2-5}}{{y}^{13-\left( -2 \right)}} \right) \\ & =-3{{x}^{-3}}{{y}^{13+2}} \\ & =-3{{x}^{-3}}{{y}^{15}} \end{align}\] Negative exponent rule: For any real number \[b\]other than zero and \[m\]a natural number \[{{b}^{-m}}=\frac{1}{{{b}^{m}}}\]. Here, \[x\ne 0\]. So, \[-3{{x}^{-3}}{{y}^{15}}=\frac{-3{{y}^{15}}}{{{x}^{3}}}\] So, \[\frac{24{{x}^{2}}{{y}^{13}}}{-8{{x}^{5}}{{y}^{-2}}}=\frac{-3{{y}^{15}}}{{{x}^{3}}}\].
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