Solution to initial condition problem Let $y(t)=-\ln(1-e^{(t+c)})$.
  I'm trying to find the solution to the initial condition $y(0)=-\ln 2$.


Isolate $c$:
$$0=\ln(2)-\ln(1-e^c)=\ln\left({2\over1-e^c}\right)$$
therefore
$$-e^c=2-1 \leftrightarrow e^c=-1 \leftrightarrow c=0$$
I can't figure out where I'm going wrong
Hint: You should rewrite the equation by replacing $e^c=a$.

$$y=-\ln(1-ae^t)$$
$c=0$ is wrong. You cannot find a real solution for $c$ because $e^c\ge 0\,$ for real $c$. A complex solution is $c=\pi i,\,$ i.e. the principal value of $\ln(-1).$