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#1 2005-12-13 13:13:45
- Chemist
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Equation
Can anyone prove the validity of this equation by deriving it :
f(AxBy) = (fA^xfB^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
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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
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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
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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.
Thank you for your time.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
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#5 2005-12-14 02:18:04
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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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