make x the subject
y=5(1-e^(-x))
first i expand it
so
y=5-5e^(-x)
loge(y)=loge(5)+loge(e5x)
loge(y)-loge(5)=5x
x=loge(y)-loge(5)/5 ??
$\displaystyle y = 5\left(1 - e^{-x}\right)$
$\displaystyle \frac{y}{5} = 1 - e^{-x}$
$\displaystyle e^{-x} = 1 - \frac{y}{5}$
$\displaystyle e^{-x} = \frac{5 - y}{5}$
$\displaystyle -x = \ln{\frac{5 - y}{5}}$
$\displaystyle x = -\ln{\frac{5 - y}{5}}$
(Anything from this step onwards is just to get the answer to whatever form you like)
$\displaystyle x = \ln{\left(\frac{5 - y}{5}\right)^{-1}}$
$\displaystyle x = \ln{\frac{5}{5 - y}}$
$\displaystyle x = \ln{5} - \ln{(5 - y)}$.