You need to substitute `1-F` for `H` in equation of `s = (1/2)(log(H/(1-H))-log(F/(1-F)))` such that:

`s = (1/2)(log((H/(1-H))-log(F/(1-F)))`

Using the equation `H = 1- F => F = 1 - H,` yields:

`s = (1/2)(log(((1-F)/(1-H))-log((1-H)/(1-F)))`

Using logarithmic identity `log(a/b) = log a - log b` yields:

`s = (1/2)(log(1-F) - log(1-H) - log(1-H) + log(1-F))`

`s = (1/2)(2log(1-F) - 2log(1-H))`

Factoring out 2 yields:

`s = log(1-F) - log(1-H)`

Since `1-F = H` , yields:

`s = logH - log(1-H)`

Converting the difference of logarithms into a quotient yields:

`s = log (H/(1-H))`

**Hence, evaluating the simplest formula for computing the measure of sensitivity s yields `s = log (H/(1-H)).` **