Answer
$$ y = \frac{1}{{1 - {e^x}}}$$
Work Step by Step
$$\eqalign{
& \ln \left( {y - 1} \right) = x + \ln y \cr
& {\text{exponentiate both sides}} \cr
& {e^{\ln \left( {y - 1} \right)}} = {e^{x + \ln y}} \cr
& {e^{\ln \left( {y - 1} \right)}} = {e^x}{e^{\ln y}} \cr
& {\text{simplifying, we get:}} \cr
& y - 1 = y{e^x} \cr
& {\text{solve for }}y \cr
& y - y{e^x} = 1 \cr
& y\left( {1 - {e^x}} \right) = 1 \cr
& y = \frac{1}{{1 - {e^x}}} \cr} $$
