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: 73

Answer

The solution set is $\{ \frac{1}{e-1}\approx 0.58 \}$.

Work Step by Step

The given equation is $\ln (x+1) -\ln x=1$ Add $\ln (x)$ to both sides. $\ln (x+1) -\ln x+\ln x=1+\ln x$ Use $1=\ln e $. $\ln (x+1) =\ln e+\ln x$ Use product rule on the right hand side. $\ln (x+1) =\ln (ex)$ $ x+1 =ex$ Isolate $x$. $x=\frac{1}{e-1}$ $x=0.58$.
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