Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.6 - Exponential and Logarithmic Equations; Further Applications - 9.6 Exercises - Page 630: 41

Answer

$x=\dfrac{1}{3}$

Work Step by Step

Since $\log_b m=\log_b n $ implies $m=n$, then the given equation, $ \log (6x+1)=\log 3 $, implies \begin{align*}\require{cancel} 6x+1&=3 .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} 6x&=3-1 \\ 6x&=2 \\\\ \dfrac{\cancel6x}{\cancel6}&=\dfrac{2}{6} \\\\ x&=\dfrac{\cancelto12}{\cancelto36} \\\\ x&=\dfrac{1}{3} .\end{align*} Hence, the solution to the equation $ \log (6x+1)=\log 3 $ is $ x=\dfrac{1}{3} $.
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