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Math Help - Please help

  1. #1
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    Please help

    Find the solution with the reasoning


    x^y=8 , y^x=9
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  2. #2
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    Quote Originally Posted by dapore View Post
    Find the solution with the reasoning


    x^y=8 , y^x=9
    hi
    x^y=2^3=8,y^x=3^2=9.
    therefore,
    x=2 and y=3.
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  3. #3
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    Thank you very much, but how to prove that this solution is the only solution
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  4. #4
    Super Member Bacterius's Avatar
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    Can you find another (integer I believe) combination of x and y that satisfy x^y = 8 ?
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  5. #5
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    Quote Originally Posted by Bacterius View Post
    Can you find another (integer I believe) combination of x and y that satisfy x^y = 8 ?
    is there anyway we can prove it,or is it just obvious.
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  6. #6
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    i was just wondering how much solutions this equation has,
    y^x-x^y=1
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  7. #7
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    Raoh
    my full gratitude and appreciation
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  8. #8
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    This my way of proof

    1 - can not be any of the X, Y number is negative because in this case it would output the left end in a fraction equations absolute value smaller than 1
    2 - can not be any of the X, Y equal to zero because in this case would be the left end result = either 1 or zero or undefined
    3 - can not be any of the X, Y equal to 1 because it has achieved one of the equations can not be achieved while the other
    And therefore there is no option for X, Y only be a Moajabin greater than 1
    4 - can not be greater than Q Ooisawi 3 because when you is no more achieved the first equation and therefore the only value for Q is 2
    5 - can not be greater than p. 4 Ooisawi because it is no more realized when the second equation and therefore the only value of r is 3

    What do you think
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