Inverse of function given by equation $y = a - b \cdot \text{ln}(1-\frac{x}{c})$ This is probably a silly question, but how do I find the inverse of the function given by the equation $y = a - b \cdot \text{ln}(1-\frac{x}{c})$? I tried doing $\text{exp()}$ on both sides but I don't know how to cope with $\text{exp}(b\cdot\text{ln}())$. Is there a simple trick I forgot about?
 A: Instead of immediately taking the exponential, I would proceed as follows.
$$
y = a - b \cdot \ln\left(1-\frac{x}{c}\right)\\
b \cdot \ln\left(1-\frac{x}{c}\right) = a-y\\
\ln\left(1-\frac{x}{c}\right) = \frac{a-y}{b}\\
1-\frac{x}{c} = \exp\left(\frac{a-y}{b}\right)\\
\frac{x}{c} = 1 - \exp\left(\frac{a-y}{b}\right)\\
x = c - c\exp\left(\frac{a-y}{b}\right)
$$

If you insist on using $\exp()$ as the first step, you could use the following simplification.
$$
\exp\left(a - b \cdot \text{ln}\left(1-\frac{x}{c}\right)\right) = \\
\frac{\exp(a)}{\exp \left(b \cdot \text{ln}\left(1-\frac{x}{c}\right)\right)} =\\
\frac{\exp(a)}{\exp \left(\text{ln}\left(1-\frac{x}{c}\right)\right)^b} =\\
\frac{\exp(a)}{\exp \left(1-\frac{x}{c}\right)^b}.
$$
A: " Is there a simple trick I forgot about?"
Yes.  If $k = m\ln n$ then $e^k = e^{m\ln n} = (e^{\ln n})^m = n^m$.
So you can do it this way (although I do not advise it)
$y = a-b\ln(1-\frac xc)$
$e^y = e^{a-b\ln(1-\frac xc)}= \frac {e^a}{e^{b\ln(1-\frac xc)}}=\frac {e^a}{(1-\frac xc)^b}$
$(1-\frac xc)^b =\frac {e^a}{e^y}= e^{a-y}$
$1-\frac xc = e^{\frac {a-y}b}$
$\frac xc = 1-e^{\frac {a-y}b}$
$x = c(1-e^{\frac {a-y}b})$.
.....
Instead I'd advice unwrapping layer by layer
$y = a-b\ln(1-\frac xc)$
$y-a = -b\ln(1-\frac xc)$
$\frac {a-y}b =\ln (1-\frac xc)$
$e^{\frac {a-y}b} = 1-\frac xc$
$\frac xc = 1-e^{\frac {a-y}b}$
$x = c(1-e^{\frac {a-y}b})$.
