Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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

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

Offline

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

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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

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

Offline

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

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

Offline

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

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

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

Werner Heisenberg

Offline