Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 7 - Exponents and Exponential Functions - 7-4 More Multiplication Properties of Exponents - Practice and Problem-Solving Exercises - Page 437: 55



Work Step by Step

Given : $(m^{2}n^{3})^{a}=\frac{1}{m^{6}n^{9}}$ (The empty square box is represented as a) This implies : $m^{2a}$ X $n^{3a}=m^{-6}$ X $n^{-9}$ (since $(ab)^{n} = a^{n}b^{n}$ and $a^{-1} = \frac{1}{a}$) Thus, equating the exponents on both sides of the equation to the corresponding bases, we get : $m^{2a}=m^{-6}$ and $n^{3a}=n^{-9}$ Hence, $2a=-6$ and $3a=-9$ This implies $a=\frac{-6}{2}=\frac{-9}{3}=-3$ Final result : The value in the blank is -3
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