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: 67



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. $\log_5x+\log_5(4x-1)=1$ $\log_5x(4x-1)=1$ $5^1=x(4x-1)$ $5=4x^2-x$ $0=4x^2-x-5$ $0=(4x-5)(x+1)$ $4x-5=0$ $4x=5$ $x=\frac{5}{4}$ $x+1=0$ $x=-1$
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