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: 54



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log(3x+5)-\log(2x+4)=0 ,$ use the properties of logarithms to simplify the left-hand expression. Then change to exponential form. Use the properties of equality to isolate the variable. Finally, do checking of the solution 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 expression above is equivalent \begin{array}{l}\require{cancel} \log\dfrac{3x+5}{2x+4}=0 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} \log_{10}\dfrac{3x+5}{2x+4}=0 \\\\ \dfrac{3x+5}{2x+4}=10^0 \\\\ \dfrac{3x+5}{2x+4}=1 .\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} (3x+5)(1)=(2x+4)(1) \\\\ 3x+5=2x+4 \\\\ 3x-2x=4-5 \\\\ x=-1 .\end{array}
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