## Calculus 8th Edition

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