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