You need to substitute 0.80 for H and 0.20 for F in equation of s to evaluate s, such that:

`s(0.80, 0.20) = (1/2)(log(H/(1-H))-log(F/(1-F))`

`s(0.80, 0.20) = (1/2)(log(0.80/(1-0.80))-log(0.20/(1-0.20))`

`s(0.80, 0.20) = (1/2)(log(0.80/0.20)-log(0.20/0.80))`

Using logarithmic identities, you may convert the logarithms of quotients in a difference of logarithms such that:

`s(0.80, 0.20) = (1/2)(log 0.80 - log 0.20 - log 0.20 + log. 0.80)`

Reducing like terms yields:

`s(0.80, 0.20) = (1/2)(2log 0.80 - 2log 0.20)`

Factoring out 2 yields:

`s(0.80, 0.20) = (2/2)(log 0.80 - log 0.20)`

Converting the difference of logarithms into the logarithm of quotient yields:

`s(0.80, 0.20) = log (0.80/0.20) => s(0.80, 0.20) = log 4`

Reasoning by analogy, you may evaluate `s(0.80, 0.60)` and `s(0.80, 0.90)` such that:

`s(0.80, 0.60) = log (0.80/0.60) => s(0.80, 0.60) = log 4/3`

`s(0.80, 0.90) = log (0.80/0.90) => s(0.80, 0.90) = log 8/9`

**Hence, evaluating `s(0.80, 0.20), s(0.80, 0.60)` and `s(0.80, 0.90)` yields `s(0.80, 0.20) = log 4 , s(0.80, 0.60) = log 4/3 ,s(0.80, 0.90) = log 8/9. ` **

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