You need to evaluate s for `(0.6,0.2), ` using the equation of s such that:

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

Substituting 0.6 for H and 0.2 for F yields:

`s = (1/2)(log(0.6/(1- 0.6)) - log(0.2/(1- 0.2)))`

`s = (1/2)(log(0.6/0.4) - log(0.2/0.8))`

Using logarithmic identities yields:

`s = (1/2)(log0.6 - log 0.4- log 0.2 + log 0.8)`

`s = (1/2)(log(3*0.2) - log0.4 - log 0.2 + log (2*0.4))`

`s = (1/2)(log 3+ log 0.2 - log 0.4 - log 0.2 + log 2 + log 0.4)`

Reducing like terms yields:

`s = (1/2)(log 3 + log 2) => s = (1/2)(log(3*2)) => s = (1/2)log 6`

You need to double s for the same value of F to evaluate H, such that:

`2s = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`2*(1/2)log 6 = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`log 6 = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`2 log 6 = log(H/(1- H)) - log(0.2/0.8)`

`log 6^2 + log (0.2/0.8) = log (H/(1- H)) => log 36*(0.2/0.8) = log (H/(1- H)) => 36*(1/4) = (H/(1- H)) => 9 = (H/(1- H))`

`9(1-H) = H => 9 - 9H = H => 9 = 10H => H = 9/10 = 0.9`

**Hence, evaluating the value of H under the given conditions yields `H = 0.9` .**

You need to double s for the same value of H to evaluate F, such that:

`log 6 = (1/2)(log(0.6/0.4) - log(F/(1- F)))`

`log 36 = log(0.6/0.4) - log(F/(1- F)) => log(F/(1- F)) = log(0.6/0.4) - log 36`

`log(F/(1- F)) = log(0.6/(36*0.4)) => log(F/(1- F)) = log(0.1/(6*0.4)) => (F/(1- F)) = 1/24 => 24F = 1 - F => 25F = 1 => F = 1/25 =0 .04`

**Hence, evaluating F under the given conditions yields `F = 0.04.` **

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