Using the $\ln(\cdot)$ for $(1-e^{-x})$ The given function: 
$$B= A(1-(e^{-x}))$$
Now, I want to 'destroy' the e-function by taking the logarithm of it.
First, since
$\ln(ab) = \ln(a) + \ln(b)$ we get that $\ln(b) = \ln(a) + \ln(1-e^{-x})$.
Using $\ln(1) =0$,  $\ln(b) = \ln(A) +x$.
Is this the right way to handle the function?
Thank you.

No it is not true that $\ln(a+b)=\ln a+\ln b$ which you used to split up $\ln(1-e^{-x})$
A better approach is to first solve for the $e^{-x}$
$$e^{-x} = 1-\frac{B}{A}$$
Now take the logarithm.
