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: 23

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: $\log_8[(x+6)(x+4)] \ ...(1)$ Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $\log (x+6)(x+4)= 8^1$ $\log (x^2+10x+24)= 8$ or, $x^2+10x+16=0$ This is a quadratic equation; thus by factoring it will become: $(x+8)(x+2)=0$ By the zero product property, we have: $x=-2$ and $x=-8$ Since the domain of the variable is $x \gt -4$, we cannot accept the value of $x=-8$ Thus, our answer is: $x= -2$
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