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: 70



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 $x$. $-6$ is not part of the domain, so $-1$ is the only answer. $\log_6(x+3)+\log_6(x+4)=1$ $\log_6(x+3)(x+4)=1$ $6^1=x^2+7x+12$ $x^2+7x+6=0$ $(x+1)(x+6)=0$ $x=-6, x=-1$ $x=-1$
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