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How to solve when x is an exponent and a base

  1. Apr 11, 2010 #1
    Just out of curiosity, I was wondering how to solve for x in an expression of the form
    ax = xb

    It kinda tripped me out at first.
  2. jcsd
  3. Apr 11, 2010 #2


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    There is no way in general to solve this kind of equation in terms of elementary functions.
  4. Apr 11, 2010 #3
    You can try manipulating it to fit the lambert W function I think.
  5. Apr 12, 2010 #4


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    Yes. Taking the "b"th root of both sides, [itex](a^{1/b})^x= x[/itex] so that [itex]1= xe^{-(1/b)(ln a)x}[/itex]. Multiply on both sides by -(1/b)ln a: [itex]-(1/b)ln a[/itex][itex]= (-(1/b)(ln a)x)e^{-(1/b)(ln a)x}[/itex].

    Letting y= -(1/b)(ln a)x, that becomes [itex]ye^y= -(1/b)ln a[/itex]. Using the Lambert W function (which is defined as the inverse function to f(x)= xex) we have y= W(-(1/b)ln a).

    Then [itex]x= -\frac{b}{ln a}W(-(1/b)ln a)[/itex]
  6. Apr 12, 2010 #5
    Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?
  7. Apr 12, 2010 #6


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    There are numerical methods, like Newton's method or the secant method, that can give numerical values for the W function -- or even your original function directly. (If you just want the W function, there are optimizations that can be used to make it go faster.)
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