Stuck solving $\ln(e^y-1)-y=t+c$ for $y$

Question

I'm trying to solve for $y$

$\ln(e^y-1)-y=t+c$

$e^y-1=e^{(t+c+y)}$

$e^y=e^{(t+c+y)}+1$

$y=t+c+y+1$

Where am I going wrong?

Answer 1

Taking from your third step $$e^y = e^{t+c+y} + 1 \\ e^y - e^{t+c+y} =1 \\ e^y(1-e^{t+c}) = 1 \\ e^y = (1-e^{t+c})^{-1} \\ y = -\ln(1-e^{t+c})$$