Answer
$7^{6}a^{2}b^{4}$
Work Step by Step
$(\frac{7^{-3}}{ab^{2}})^{-2}$
Using $\frac{1}{a^{n}}= a^{-n}$
$= (7^{-3}a^{-1}b^{-2})^{-2}$
$= (7^{-3})^{-2}(a^{-1})^{-2}(b^{-2})^{-2}$
Using $(a^{m})^{n} =a^{m \times n} $
$=7^{6}a^{2}b^{4}$
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321785045
ISBN 13:
978-0-32178-504-6
Chapter 5 - Section 5.2 - More Work with Exponents and Scientific Notation - Exercise Set - Page 268: 38
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