College Algebra (11th Edition)

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

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 60



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_2(5x-6)-\log_2(x+1)=\log_23 ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use the properties of equality and concepts on solving quadratic equations. 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} \log_2\dfrac{5x-6}{x+1}=\log_23 .\end{array} Since the logarithm on both sides have the same base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5x-6}{x+1}=3 .\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{5x-6}{x+1}=\dfrac{3}{1} \\\\ (5x-6)(1)=(x+1)(3) \\\\ 5x-6=3x+3 \\\\ 5x-3x=3+6 \\\\ 2x=9 \\\\ x=\dfrac{9}{2} .\end{array} Upon checking, $ x=\dfrac{9}{2} $ satisfies the original equation.
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