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: 68



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 $x$. The negative answer is not part of the domain. $\log_6(x+5)+\log_6x=2$ $\log_6x(x+5)=2$ $36=x(x+5)$ $0=x^2+5x-36$ $0=(x+9)(x-4)$ $x=-9, x=4$ $x=4$
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