Algebra and Trigonometry 10th Edition

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: 58



Work Step by Step

$\ln(x+1)-\ln(x-2)=\ln x$ $\ln\frac{x+1}{x-2}=\ln x$ Using the One-to-One Property: $\frac{x+1}{x-2}=x$ $x+1=x^2-2x$ $0=x^2-3x-1~~$ ($a=1,b=-3,c=-1$): $x=\frac{-b±\sqrt {b^2-4ac}}{2a}=\frac{-(-3)±\sqrt {(-3)^2-4(1)(-1)}}{2(1)}=\frac{3±\sqrt {13}}{2}$ $x=\frac{3+\sqrt {13}}{2}=3.303$ $x=\frac{3-\sqrt {13}}{2}=-0.303$. But since $-0.303$ is not in the domain of $\ln x$, it is not a valid solution.
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