x^y=y^x

Discussion in 'Physics & Math' started by BloodSuckingGerbile, Jul 5, 2002.

  1. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    x^y=y^x

    x,y are whole numbers (Z).

    Can you solve it?
     
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  3. James R Just this guy, you know? Staff Member

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    x and y are any two integers such that x=y.
     
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  5. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    How exactly did you get to the answer (mathematically)?
     
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  7. itchy Registered Senior Member

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    2^4 = 4^2
     
  8. Merlijn curious cat Registered Senior Member

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    My first reaction is x=y.
    However, a more unified answer is harder to find.
    this is developing as I type... so maybe I'll make a mistake or get stuck (forgive me if I do

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    )
    let y=x+a
    x^y = y^x <=>

    x^(x+a) = (x+a)^x <=>
    x^x + x^a = x^x + x*ax^(x-1) + ...+x*xa^(x-1)+a^x (Pascal's sequence) <=>
    x^a = x*ax^(x-1)+ 2(x-1)*a^2*x^(x-2)+...+2(x-1)*a(x-2)*x^2+x*xa^(x-1)+a^x <=>
    some cleaning up to do...
    x^a = ax^x + ... Hmmm I need a paper now. I'll get back to this later.
     
  9. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    Merlijn, you're a genious.
     
  10. Merlijn curious cat Registered Senior Member

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    does this mean you have the answer now?
    or do I have to continue?
    (or were you merely sarcastic?)

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  11. James R Just this guy, you know? Staff Member

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    BloodSuckingGerbile:

    What's <b>your</b> answer to this?
     
  12. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    440
    No........
    I don't have the answer... yet.
    But you gave me a starting point.
    continue, of course!
    Let's see who gets first to the answer.

    James R:
    I don't know the answer. Trying to find it, though.

    I took Marlijns' example and tried doing what he did but it seemed too much to "clean up" so I assumed y=ax

    and I got

    x^(ax)=(xa)^x
    ||
    \/
    (x^x)^a=x^x*a^x
    ||
    \/
    (x^x)^(a-1)=a^x

    And then I did

    ln[(x^x)^(a-1)]=ln(a^x)
    || ||
    (a-1)*x*lnx = x*lna
    ||
    \/
    lnx=(lna)/(a-1)

    and then

    e^(lnx)=e^[(lna)/a-1)]
    || ||
    x = (a-1)st root of a = a^[1/(a-1)]

    try it on a calculator. It actually works. Now, this is true for real numbers, but I need to find all a's for which the expression

    a^[1/(a-1)] = Z

    is true.

    Merlijn, try to finish what you started. Maybe that will give the answer.
     
  13. Merlijn curious cat Registered Senior Member

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    1,014
    BloodSuckingGerbile,
    Ok, I will try to work on it; it's a lot of cleaning up though. Maybe there is an easier way.
    However, I seriously doubt that y=ax makes it any easier.

    bye
     
  14. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    440
    I think I have the answer.

    (a-1)*x*lnx = x*lna

    (before I divided lna by a-1)
    Gives the answer a=1, or y=x*1=x, which is true.
    Then when I divided lna by (a-1)
    I got

    lnx=(lna)/(a-1)

    and then

    x=a^[1/(a-1)]

    So there is only one integer a for which a^[1/(a-1)] is an integer and that is a=2, which gives x=2 and y=a*x=2*2=4.

    :bugeye:

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    Just one more thing: -2 and -4 are also true...

    %@#$%!!!!!!

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    I guess I'll have to do that from the beginning...
     
  15. Merlijn curious cat Registered Senior Member

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    well you know that with your
    y=ax
    there is a solution a=1 (for all x) and a=2 (at least for x=2)
    negative solutions for a are higly improbable (hmmm x=1, a=-1 gives 1/1 = -1 ? eeeh obviously not!

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    )
    Forget the negative solutions.
     
  16. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    But the negative solutions exist.
    Merlijn, you think that with y=x+a you can get them too?
     
  17. Merlijn curious cat Registered Senior Member

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    1,014
    well only if both x and y are negative... but not only one of the two...
    I haven't had time yet to have another look at the problem... be patient pleace.
    (I am sure there are others who can solve the problem)
     
  18. (Q) Encephaloid Martini Valued Senior Member

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    x=2
    y=4
     
  19. overdoze human Registered Senior Member

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    Cool problem, gerbil

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    I think your approach works well, but you have to take care. As such:

    y = ax, x^y = y^x

    x^(ax) = (ax)^x

    (x^a)^x = (xa)^x

    But this is where we have to pay attention. If x is odd, then we can just raise both sides to power 1/x to obtain:

    x^a = xa, x odd

    If x is even then we have two possibilities:

    x^a = xa, and x^a = -xa.

    In either case, if a < 0 then |1/x^|a|| = |xa| => 1/|a| = |x^(|a|+1)|. Since x and a are integers, 1/|a| must be an integer which can only be if |a|=1, but a is even. Therefore a>0. Note that this applies regardless of whether x is odd or even (which means y and x must have the same sign.)

    In the second case, let's consider the possible values of a:

    a=2: x^2 = -2x => x = -2, y = -4
    a=4: x^4 = -4x => x^3 = -4 => -1 < x < -2
    a=6: x^6 = -6x => x^5 = -6 => -1 < x < -2
    ...
    as a approaches infinity, x approaches -1 but is never again an integer. Which means the only solution for this case is x=-2,y=-4.

    Now let's return to the other case:

    x^a = xa, a>0

    So let's try some values of a:

    a=1: x = x => y = x
    a=2: x^2 = 2x => x = 2, y = 4
    a=3: x^3 = 3x => x^2 = 3 => 1 < x < 2
    a=4: x^4 = 4x => x^3 = 4 => 1 < x < 2
    ...
    In the limit, as a becomes infinite x approaches 1 but is never again an integer. Which means that all values of a greater than 2 are excluded.

    So your solution is:

    a=1: y=x, x any integer (positive or negative)
    a=2: x=2,y=4 and x=-2,y=-4

    That's it as far as integer solutions go, unless I missed something...
     
  20. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    Overdoze, thanks for fully answering my question.

    Not being careful is my biggest problem.

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  21. Han Baumer Member Registered Senior Member

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    I was just playing around with Mathematica to see what it can do.
    It came up with the solution:

    -x * ProductLog[ -Log[x]/x ]
    ________________________
    Log[x]

    This is an interesting function: for 0<=x<=e it evaluates to x, at e it has its maximum, after which it decreases and has limit 1. So the only Integer solutions this function gives at y=2 are x=2 and x=4. Other integer solutions are x=y=0 and x=y=1.
    GRAPH

    Why Mathematica misses the obvious x=y solution is a mystery to me. Maybe I don't know the program well enough.
     
  22. BloodSuckingGerbile Master of Puppets Registered Senior Member

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    hmm...
    That sounds interesting.
    I've never heard of Mathematica.
    Some kind of software?
    Tell me more, please.
     
  23. Han Baumer Member Registered Senior Member

    Messages:
    41
    Mathematica is the program from Wolfram Research that can do symbolic mathematics. See http://www.wolfram.com

    Greetings,

    Han.
     

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