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## #1 2005-12-13 13:13:45

Chemist
Member
Registered: 2005-12-12
Posts: 35

### Equation

Can anyone prove the validity of this equation by deriving it :

f(AxBy) = (fA^xfB^y)^[1/(x+y)]  ?

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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## #2 2005-12-13 15:57:15

ryos
Member
Registered: 2005-08-04
Posts: 394

### Re: Equation

a^x*a^y = a^(x+y), but it only works for a common base. Therefore,

ƒ(AxBy) = ƒ(Ax) * ƒ(By) = [ƒ(Ax) * ƒ(By)]^[(x+y)(1/x+y)] = [ƒ(Ax)^x * ƒ(By)^y]^[1/(x+y)]
if and only if ƒ(Ax) = ƒ(By).

Last edited by ryos (2005-12-13 16:05:53)

El que pega primero pega dos veces.

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## #3 2005-12-13 19:51:45

MathsIsFun
Registered: 2005-01-21
Posts: 7,664

### Re: Equation

Ahhh... bases! A is in base x, and B is in base y, right?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #4 2005-12-14 02:11:45

Chemist
Member
Registered: 2005-12-12
Posts: 35

### Re: Equation

My post wasn't very clear, so let me first clarify that it is not given that f(Ax) = f(By). I should add that f denotes an activity coefficient, so it's a constant.

And yes MathsIsFun , A is in base x, and B is in base y. Hope you figure it out cos I've been told it's a difficult problem.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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## #5 2005-12-14 02:18:04

Chemist
Member
Registered: 2005-12-12
Posts: 35

### Re: Equation

Call f anything you wish k, n, l, ...  it is not  , a function.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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