College Algebra (11th Edition)

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

Chapter 4 - Test - Page 472: 21



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_3(x+1)-\log_3(x-3)=2 ,$ use the properties of logarithms to simplify the expression at the left. Then convert to exponential form and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_3 \dfrac{x+1}{x-3}=2 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \dfrac{x+1}{x-3}=3^2 \\\\ \dfrac{x+1}{x-3}=9 .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{x+1}{x-3}=\dfrac{9}{1} \\\\ (x+1)(1)=(x-3)(9) \\\\ x+1=9x-27 \\\\ x-9x=-27-1 \\\\ -8x=-28 \\\\ x=\dfrac{-28}{-8} \\\\ x=\dfrac{7}{2} .\end{array}
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