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: 2

Answer

$\frac{1}{16}$

Work Step by Step

We are given the expression $(2^{-2})^{2}$. To simplify, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $(2^{-2})^{2}=2^{-2\times2}=2^{-4}$ To simplify this into a positive exponent, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer). Therefore, $2^{-4}=\frac{1}{2^{4}}=\frac{1}{2\times2\times2\times2}=\frac{1}{16}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.