College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises - Page 469: 63



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \ln(6x)-\ln(x+1)=\ln4 ,$ use the laws of logarithms to simplify the expression. Then drop the logarithm. Use properties of equality to isolate the variable. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \ln\dfrac{6x}{x+1}=\ln4 .\end{array} Since both sides have the same logarithmic base, then the logarithm may be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{6x}{x+1}=4 .\end{array} Using cross-multiplication and the Distributive Property, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{6x}{x+1}=\dfrac{4}{1} \\\\ 6x(1)=(x+1)(4) \\\\ 6x=x(4)+1(4) \\\\ 6x=4x+4 \\\\ 6x-4x=4 \\\\ 2x=4 \\\\ x=\dfrac{4}{2} \\\\ x=2 .\end{array}
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