Answer
$x=\frac{-1±\sqrt (1+8e^{2})}{4}$
Work Step by Step
Given: $ln(2x+1)=2-lnx$
$(2x+1)=e^{2-lnx}$
$(2x+1)=\frac{e^{2}}{x}$
$2x^{2}+x-e^{2}=0$
This implies
$x=\frac{-1±\sqrt (1+8e^{2})}{4}$
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