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

$\frac{-3{{y}^{15}}}{{{x}^{3}}}$.

#### 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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