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.

You seem to have accidentally used the incorrect formula $\ln(a+b)=\ln(a)+\ln(b)$. This is confusing, because the distributive property $c(a+b)=ca+cb$ looks so similar.
Someone once asked a student why she had accidentally assumed that $(a+b)^2=a^2+b^2$. The answer: "I thought we could do that any time we had parentheses." So it's a good idea to emphasize: the distributive property is a property of the multiplication, not parentheses! The confusion could come from the fact that multiplication isn't written out (as we write $ab$ rather than $a\times b$). The distributive property says that:
$$c\times(a+b)=(c\times a)+(c\times b)$$
(Some of those parentheses are redundant.) Note that it only talks about how multiplication behaves with addition; it does not mention anything about exponents or logs. The parentheses are just there to mark what operations are done first, nothing else.
(It's a good idea for you to try to prove the distributive property and see why it's true.)
