>>13673285I made a mistake at the end. The 0.632 is wrogn and unnecessary but it somehow compensated for the fact that the original function:
>y= y0 + ln(1-(x/x0))evaluates to y=0 for x=x0*(1 - 1/exp(y0)). this means that for a y0 reasonably larger than 1 your function will evaluate to 0 very close to specified x0 (since 1/exp(y0) will be negligible).
however if you want to use that for small numbers or fractions, another term needs to be added:
>y= y0 + k*ln(1-(x/x0))where k = 4/y0. this is due to a property of the original inverse exponential called the time constant. once you add that k-term the function is all good in terms of x0 and y0
