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Solve an exponent equation without restricting the domain?

  1. Jun 10, 2012 #1
    It's a pretty basic equation:

    x = x^u, solve for u.

    ln x = u ln x

    ln x / ln x = u

    u = 1

    However, this solution only applies for x > 0. Is there any way we could solve this equation for all values of x?
     
  2. jcsd
  3. Jun 11, 2012 #2
    The equation has infinitely many solutions when x = 0, so you're going to have to partition the domain anyway. For negative values of x specifically, you might try toying with the inverse of xu...
     
  4. Jun 11, 2012 #3

    chiro

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    Hey Hertz and welcome to the forums.

    Are you are aware of complex numbers and Eulers identity?
     
  5. Jun 11, 2012 #4
    Thanks :)

    I do know about complex numbers, but not Eulers identity. I am more than happy to do research though to solve this problem.
     
  6. Jun 11, 2012 #5

    haruspex

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    I'm not sure that xu is defined on the reals when x negative. If taking x and u to be complex then try substituting x = r.e, r ≥ 0.
     
  7. Jun 11, 2012 #6
    I'm not 100% sure if this situation is case specific, but using Euler's Identity I think I've proven it for u < 0 also.

    Euler's Identity: e^ipi + 1 = 0

    e^ipi = -1
    ln(-1) = ipi

    Therefore:
    ln(-a) = ln(-1) + ln(a)
    ln(-a) = ipi + ln(a)

    x^u = x
    u ln x = ln x
    u = ln x/ln x

    Even if x is negative, the ln x's still cancel out.

    so, u = 1 for all x such that x != 0.

    Thank you for the suggestion chiro.

    One final quick question; using a Maclaurin or Taylor series would it be possible to evaluate ln x when x = 0? Sorry if this question sounds foolish, I haven't learned much about them yet.
     
  8. Jun 11, 2012 #7
    The series would diverge, as the logarithm would.
     
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