Answer
$x=-1.156$
Work Step by Step
$\ln x-\ln(x+1)=2$
$\ln\frac{x}{x+1}=2$
$e^{\ln\frac{x}{x+1}}=e^2$
$\frac{x}{x+1}=e^2$
$x=e^2x+e^2$
$x(1-e^2)=e^2$
$x=\frac{e^2}{1-e^2}=-1.156$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 395: 55
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