Answer
$j^{32}/32k^{26}$
Work Step by Step
$(2j^2k^4)^{-5}*(k^{-1}j^7)^{6}$
$2^{-5}*j^{-10}*k^{-20}*k^{-6}*j^{42}$
$1/32*j^{32}*k^{-26}$
$j^{32}/32k^{26}$
Algebra 1
by Hall, Prentice
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: 36
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