Answer
$\frac{1}{e^2-1}\approx0.157$
Work Step by Step
Step 1. Rewrite the equation as $ln(x+1)=ln(e^2)+ln(x) \longrightarrow ln(x+1)=ln(xe^2)$
Step 2. Thus $x+1=xe^2\longrightarrow (e^2-1)x=1\longrightarrow x=\frac{1}{e^2-1}$.
Step 3. Check answers, $x=\frac{1}{e^2-1}\approx0.157$ fits the original equation.
