Answer
$$y=\frac{1}{1-e^x}$$
Work Step by Step
$$\ln(y-1)=x+\ln y$$
We can rewrite $x=\ln e^x$
$$\ln(y-1)=\ln e^x+\ln y$$
Also, $\ln x+\ln y=\ln(xy)$
$$\ln(y-1)=\ln(e^xy)$$
We can take out the natural logarithm on both sides.
$$y-1=e^xy$$ $$y-e^xy=1$$ $$y(1-e^x)=1$$ $$y=\frac{1}{1-e^x}$$
