Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 9 - Section 9.5 - Exponential and Logarithmic Equations - Exercise Set - Page 727: 71

Answer

No solution or $\varnothing $.

Work Step by Step

The given equation is $\log (3x-5) -\log (5x)=2$ Use quotient rule. $\log \left [\frac{3x-5}{5x} \right]=2$ Rewrite the common logarithm showing base $10$. $\log_{10} \left [\frac{3x-5}{5x} \right]=2$ Rewrite in exponential form. $\frac{3x-5}{5x}= 10^2$ Multiply both sides by $5x$. $(5x)\cdot \frac{3x-5}{5x}= (5x)\cdot 100$ Clear the parentheses. $3x-5=500x$ Add $-3x$ to both sides. $3x-5-3x=500x-3x$ Simplify. $-5=497x$ Divide both sides by $497$. $-\frac{5}{497}=\frac{497x}{497}$ Simplify. $-\frac{5}{497}=x$ Negative values are not defined for log functions.
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