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 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 337: 12



Work Step by Step

The domain of the variable requires that $x>0$. Recall: $\log_a M^r = r \log_a M$ Using the rule above gives: $-2 \log_4 x = \log_4 x^{-2}$ Thus, the given equation is equivalent to: $\log_4{\left(x^{-2}\right)} = \log_4 9$ Recall also that: $\text{If } \log_a M = \log_a N \text{, then } M=N$ Therefore, $x^{-2} = 9$ $\left(x^{-2}\right)^{-1} = (9)^{-1}$ $x^2 = \dfrac{1}{9}$ $x = \pm \sqrt{\dfrac{1}{9}}$ $x = \pm \dfrac{1}{3}$ $x$ cannot be negative so $x=-\frac{1}{3}$ will be rejected. Hence, $x= \boxed{\dfrac{1}{3}}$
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.