Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.3 Exponential Functions - 4.3 Assess Your Understanding - Page 307: 81



Work Step by Step

In order to solve the given problem, we will use the following two rules as follows: $ (a) a^{-p}=\dfrac{1}{a^p} \\ (b) a^{pq}=(a^p)^q$ We will use rule-(a) as: $4^{-2x}=\dfrac{1}{4^{2x}}$ Now, we will use rule-(b) as: $4^{-2x}=\dfrac{1}{(4^{x})^2}$ When $4^x=7$, then we simplify the expression as: $4^{-x}=\dfrac{1}{7}$ Therefore, $4^{-2x}=(4^{-x})^{2} =(\dfrac{1}{7})^2=\dfrac{1}{49}$
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.