Inverse of function given by equation $y = a - b \cdot \text{ln}(1-\frac{x}{c})$

Solution 1:

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. $$ \exp\left(a - b \cdot \text{ln}\left(1-\frac{x}{c}\right)\right) = \\ \frac{\exp(a)}{\exp \left(b \cdot \text{ln}\left(1-\frac{x}{c}\right)\right)} =\\ \frac{\exp(a)}{\exp \left(\text{ln}\left(1-\frac{x}{c}\right)\right)^b} =\\ \frac{\exp(a)}{\exp \left(1-\frac{x}{c}\right)^b}. $$

Solution 2:

" 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

$y = a-b\ln(1-\frac xc)$

$y-a = -b\ln(1-\frac xc)$

$\frac {a-y}b =\ln (1-\frac xc)$

$e^{\frac {a-y}b} = 1-\frac xc$

$\frac xc = 1-e^{\frac {a-y}b}$

$x = c(1-e^{\frac {a-y}b})$.