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Solving a polynomial of arbitrary powers

  1. Jul 5, 2007 #1
    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.

    Last edited: Jul 5, 2007
  2. jcsd
  3. Jul 6, 2007 #2


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    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.
  4. Jul 6, 2007 #3
    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.

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