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: 63


$x \approx 0.861$

Work Step by Step

Re-write the first term as: $25^x=(5^{2})^{x}=(5^{x})^{2}$ We wish to substitute $a=5^{x}$, so the equation becomes: $a^{2}-8a+16=0 $ This gives a quadratic equation whose factors are: $(t-4)^2=0$ Use the zero factor property to obtain: $ t-4 =0 \implies t=4$ Therefore, $5^{x}=4$ or, $x= \log_5 4 \approx 0.861$ Thus, our answer is: $x \approx 0.861$
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