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Solve the equation

  1. Oct 21, 2015 #1
    a<b, a^b=b^a, how to solve such kind of equation?
     
  2. jcsd
  3. Oct 21, 2015 #2

    andrewkirk

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    I doubt there's a general technique. Why not try a few small integers and see what happens though?
     
  4. Oct 21, 2015 #3

    andrewkirk

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    I take it back. There is a technique. Start by taking logs.
     
  5. Oct 21, 2015 #4
    What exactly do you mean by 'solve'? Even by taking log, i think 'a' will always exist in terms of 'b' and vice versa.
     
  6. Oct 22, 2015 #5
    Hmmm. I don't see a general technique right away... obviously if a=b, then the solution is 1, but if a<b...

    To check out the boundaries of the trichotomy of reals:

    ## (-2)^{-1} = (-1)^{-2} \rightarrow -\frac{1}{2} = 1 ##
    ## (-1)^{0} = (0)^{-1} \rightarrow 1 = 0 ##
    ## (0)^{1} = (1)^{0} \rightarrow 0 = 1 ##
    ## (1)^{2} = (2)^{1} \rightarrow 1 = 2 ##

    To check out large gaps:
    ## (2)^{2000} = (2000)^{2} \rightarrow big = small ##
    ## (-2)^{-2000} = (-2000)^{-2} \rightarrow small = big ##

    Take the b-th root of both sides:
    ## a = b^{\frac{a}{b}} ##

    Take the log base a of both sides and then exponentiate on the value a:
    ## b = a \cdot log_a b ##
    ## a^b = a^a \cdot b ##

    Set the last with the first on ## a^b ## to get:
    ## a^a = b^{b-1} ##

    Yeah, I'm thinking that this doesn't violate what I thought was a rule that it's not possible to solve one equation with two unknowns.
     
  7. Oct 22, 2015 #6

    andrewkirk

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    It can't be solved in the sense of identifying a unique solution (a,b). But it can be solved in the sense of finding a range for a for which a solution exists, and the corresponding b can then be found for any a in that range. I think finding the value of b would require numerical techniques as I don't think there's an analytic solution.

    Having said that, there's a neat pair of small positive integers that is one solution.
     
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