Solving a polynomial of arbitrary powers

1. Jul 5, 2007

davee123

W*C^(Z/A) = X*C^(Y/A) - D*X + D*W

How do you go about solving for A?
Known: C>1
Also, if it makes it easier (it shouldn't) you can assume that D=1.

If Y and Z are known, I figure you can substitute k=C^(1/A), which turns it into a more trivial polynomial. But if Y and Z *aren't* known, I'm at a loss.

For the sake of reference, here's where the problem comes from:

I want a logarithmic function that I can use in a program I'm writing. Therefore, I'm guessing that I want an equasion of the form:

f(x) = A*log(B*x + D) (that's log base C)

And I know that:
f(0) = 0
f(W) = Y
f(X) = Z

Since F(0) = 0, I can assume that D = 1. No problem.
And I don't really care what C is, nor do I think I ought to-- if it matters, I'm just assuming that C = e.

Plugging in W and solving for B in terms of A gives me:

B = (C^(Y/A) - D)/W

Plugging in X and solving for A gives me something I can't solve-- at least not in the general case:

W*C^(Z/A) = X*C^(Y/A) - D*X + D*W

And if I do the reverse, and try solving for B instead of A, I get:

(B*W+D)^Z = (B*X+D)^Y

Which I similarly can't figure out how to solve for B.
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DaveE

Last edited: Jul 5, 2007
2. Jul 6, 2007

HallsofIvy

First, that's not a polynomial. Second, since it has nothing to do with "Linear and Abstract Algebra", I am moving this to "General Math".

I don't know why you would say that. You can solve FOR A in terms of the other constants. If you let k= C1/A you get
WkZ= XkY- DX+ DY

Now, if Z and Y are positive integers, that IS a polynomial. However, there are no general formulas for solving polynomials.

3. Jul 6, 2007

davee123

D'oh! Yeah, I wasn't sure-- I figured I'd look for an "Algebra" forum, but that didn't exist.

Ahh, ok. I never realized polynomials required positive integers. Hm. I suppose as long as Z and Y are rational (and both positive or both negative), you could choose a better substitution that still turns it into a polynomial, but of course, there's still no guarantee.

DaveE