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Larson 3.3.25

  1. Jan 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine all positive rational solutions of x^y=y^x.

    2. Relevant equations

    3. The attempt at a solution
    Obviously, x=y will always work. I think that is the only solution. If I can show that x^y must be rational, I think it will be easy because then both x and y must have the same primes in the numerator and the denominator. I tried writing out x=r/s, y = t/u, and manipulating, which leads to
    [tex]r^{ts} u^{ru} = s^{ts} t^{ru} [/tex]
    which is really not helpful.
  2. jcsd
  3. Jan 4, 2008 #2


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    How about x = 2 , y = 4 ?
  4. Jan 4, 2008 #3

    Shooting Star

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    > I think that is the only solution.

    What about 2^4 = 4^2...
  5. Jan 5, 2008 #4
    Hmmm. Maybe that is the only exception. But can we prove that we have found them all....
  6. Jan 5, 2008 #5


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    This has been discussed in these forums a few times before, but I can't seem to find the threads. I did find this however.
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