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: 24

Answer

$x= 2$

Work Step by Step

Apply the logarithmic property: $\log_a M+\log_a N = \log_a (MN)$ and rearrange the given expression to obtain: $\log_5[(x+3)(x-1)] \ ...(1)$ Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $\log (x+3)(x-1)= 5^1$ $\log (x^2+2x-3)=5$ or, $x^2+2x-8=0$ This is a quadratic equation; thus by factoring, it will become: $(x+4)(x-2)=0$ By the zero product property, we have: $x=2$ and $x=-4$ Since the domain of the variable is $x \gt 0$, we cannot accept the value of $x=-4$ Thus, our answer is: $x= 2$
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