Intermediate Algebra (6th Edition)

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: 28

Answer

$\dfrac{1}{z^{26}} $

Work Step by Step

Using laws of exponents, the given expression simplifies to \begin{align*} & \left( z^{-2} \right)^{13} \\&= z^{-2(13)} \\&= z^{-26} \\\\&= \dfrac{1}{z^{26}} .\end{align*}
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