Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 101



Work Step by Step

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{1}{4x})^{-2}=\frac{1}{(\frac{1}{4x})^{2}}=(\frac{4x}{1})^{2}=\frac{4^{2}x^{2}}{1^{2}}=\frac{16x^{2}}{1}=16x^{2}$.
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.