College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 69



Work Step by Step

Use the definition of the log function to solve for $x$. Rewrite the problem so that the base has the exponent of the opposite side and is equal to the log's number and/or variable. Then simplify the side with $x$ by factoring and then solving for $x$. $-7$ is not part of the domain, so $-3$ is the only answer. $\log_3(x+6)+\log_3(x+4)=1$ $\log_3(x+6)(x+4)=1$ $3^1=x^2+10x+24$ $x^2=10x+21=0$ $(x+7)(x+3)=0$ $x=-7, x=-3$ $x=-3$
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