You need to prove that `a = 10 s` , if `s = (1/2) (log(H/(1-H)) - log(F/(1 - F)))` , hence, using logarithmic identities yields:

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

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

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

Converting the power into a square root yields:

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

You need to equate 10 s and a such that:

`a = 10 s => a = 10 log (sqrt((H(1 - F))/(F(1 - H)))) => a = log (((H(1 - F))/(F(1 - H)))^(10/2)) `

`a = log (((H(1 - F))/(F(1 - H)))^5) `

**Hence, evaluating the value of a, under the given conditions, yields that `a = log (((H(1 - F))/(F(1 - H)))^5)` and the indicated result `a = (sqrt((H(1 - F))/(F(1 - H)))) ` does not hold.**

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now