1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook