Yes, you use the fact that $\ln u=v\iff u=e^v.$ So applying this to what you have (with $u={x+1\over x-1}$ and $v=2$): $ \tag{1}{x+1\over x-1}=e^2. $ Multiplying both sides by $x-1$ gives $ \tag{2}x+1=e^2(x-1). $ (note, here that $x=1$ is not a solution of (2); so (1) and (2) are equivalent equations)
Solving for $x$ in (2):
$\eqalign{ &x+1= e^2 x-e^2\cr &\iff x-e^2x =-1-e^2\cr &\iff x(1-e^2)=-1-e^2\cr &\iff x= {-1-e^2\over 1-e^2} } $ Or $ x={e^2+1\over e^2-1}. $
When solving logarithmic equations, you sould always check your answers. In particular, check that you don't wind up taking the logarithm of a non-positive quantity.