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: 67


$x \approx 1.161$

Work Step by Step

We wish to substitute $t=4^{x}$, so the equation becomes: $a -(10)(\dfrac{1}{t}) -3=0 \implies a^{2}-3t-10=0 $ This gives a quadratic equation, whose factors are: $(t-5)(t+2)=0$ Use the zero factor property to obtain: $ t-5 =0 \implies t=5$ and $t+2=0 \implies t=-2$ But $t=4^{x}=-2$ cannot be a solution as $4^{x}$ can never be negative. So, we will consider $t=4^{x}=5$ Therefore, $4^{x}=5$ or, $x= \log_4 5 \approx 1.161$ Thus, our answer is: $x \approx 1.161$
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.