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: 22


$x= -1$

Work Step by Step

Apply the logarithmic property: $\log_a M+\log_a N = \log_a (MN)$ and rearrange the given expression to obtain: $\log_6[(x+4)(x+3)] \ ...(1)$ Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$, then we have: $\log (x+4)(x+3)= 6^1$ $\log (x^2+7x+12)= 6$ or, $x^2+7x+6=0$ This is a quadratic equation; thus by factoring, it will become: $(x+6)(x+1)=0$ By the zero product property, we have: $x=-6$ and $x=-1$ Since, the domain of the variable is $x \gt -3$, we cannot accept the value of $x=-6$ Thus, our answer is: $x= -1$
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