College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises: 66

Answer

$x= \frac{-1+\sqrt{9+4e}}{2}\approx 1.7290$

Work Step by Step

$\ln(x-1)+\ln(x+2)=1$ $\ln[(x-1)(x+2)]=1$ $(x-1)(x+2)=e^1$ $x^{2}+x-2=e^1$ $x^{2}+x-2-e^1=0$ $x^{2}+x-(2+e)=0$ We use the quadratic formula: $x= \frac{-1\pm\sqrt{1^2-4(1)(-2-e)}}{2*1}=\frac{-1\pm\sqrt{9+4e}}{2}$ However, $x= \frac{-1-\sqrt{9+4e}}{2}$ is negative and produces a negative log in the original equation (undefined). So we throw this solution out. Therefore, the only solution is: $x= \frac{-1+\sqrt{9+4e}}{2}\approx 1.7290$
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