## University Calculus: Early Transcendentals (3rd Edition)

$$y=\frac{1}{1-e^x}$$
$$\ln(y-1)=x+\ln y$$ 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. $$y-1=e^xy$$ $$y-e^xy=1$$ $$y(1-e^x)=1$$ $$y=\frac{1}{1-e^x}$$