College Algebra (11th Edition)

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

Chapter 4 - Review Exercises: 62

Answer

$x=4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_7(3x+2)-\log_7(x-2)=1 ,$ use the laws of logarithms to simplify the expression. Then change to exponential form. 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} \log_7\dfrac{3x+2}{x-2}=1 .\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{3x+2}{x-2}=7^1 \\\\ \dfrac{3x+2}{x-2}=7 .\end{array} Using cross-multiplication and the Distributive Property, the expression above is equivalent to \begin{array}{l}\require{cancel} (3x+2)(1)=(x-2)(7) \\\\ 3x+2=x(7)-2(7) \\\\ 3x+2=7x-14 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 3x+2=7x-14 \\\\ 3x-7x=-14-2 \\\\ -4x=-16 \\\\ x=\dfrac{-16}{-4} \\\\ x=4 .\end{array}
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