Answer
$x=\dfrac{1}{e^2-1} \approx 0.157$
Work Step by Step
Apply the logarithmic property:
$\log_a(\dfrac{M}{N}) = \log_a M-\log_a N$ and rearrange the given expression to obtain:
$\ln \left(\dfrac{x+1}{x} \right) = 2$
or, $e^2 =\dfrac{x+1}{x}$
or, $e^2 \ x=x+1$
or, $x(e^2-1)=1$
or, $x=\dfrac{1}{e^2-1}$
Thus, our answer is: $x=\dfrac{1}{e^2-1} \approx 0.157$
