Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3 Logarithmic Functions - 6.3 Exercises - Page 427: 36


$\frac{-1+\sqrt {({1+8e^{2}})}}{4}$

Work Step by Step

Use logarithmic properties, $logx+logy=log(xy)$ $ln(2x+1)=2-lnx$ $ln(2x+1)+lnx=2$ $log[(2x+1)x]=2$ $[(2x+1)x]=e^{2}$ $2x^{2}+x-e^{2}=0$ Use quadratic formula to find its roots. $x=\frac{-1±\sqrt {({1+8e^{2}})}}{4}$ Neglect negative root. Therefore, $x=\frac{-1+\sqrt {({1+8e^{2}})}}{4}$
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