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: 68

Answer

$x=4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \ln(5+4x)-\ln(3+x)=\ln3 ,$ use the properties of logarithms to combine the logarithmic expressions. 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 to \begin{array}{l}\require{cancel} \ln\dfrac{5+4x}{3+x}=\ln3 .\end{array} Since the logarithmic expressions on both sides have the same base, then the logarithm may be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{5+4x}{3+x}=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{5+4x}{3+x}=\dfrac{3}{1} \\\\ (5+4x)(1)=(3+x)(3) \\\\ 5+4x=9+3x \\\\ 4x-3x=9-5 \\\\ x=4 .\end{array} Upon checking, $ x=4 $ satisfies the original equation.
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