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: 64


$x \approx 0.613$

Work Step by Step

Re-write the first term as: $36^x=(6^{2})^{x}=(6^{x})^{2}$ We wish to substitute $a=6^{x}$, so the equation becomes: $a^{2}-6a+9=0 $ This gives a quadratic equation, whose factors are: $(t-3)^2=0$ Use the zero factor property to obtain: $ t- 3 =0 \implies t= 3$ Therefore, $6^{x}=3$ or, $x= \log_6 3 \approx 0.613$ Thus, our answer is: $x \approx 0.613$
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