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Math Help - Why this inequation?

  1. #1
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    Why this inequation?

    Given that A^X + B^Y = C^Z, where A, B, C > 0 and X, Y, Z >2
    what is the reason that Z < 2*max(XlogA, YlogB)??
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chauhan View Post
    Given that A^X + B^Y = C^Z, where A, B, C > 0 and X, Y, Z >2
    what is the reason that Z < 2*max(XlogA, YlogB)??
    WLOG  x\log a>y\log b and suppose  z=2x\log_c a

     a^x+b^y=c^{x\log_c a}+c^{y\log_c b}<2c^{x\log_c a}\leq c^{2x\log_c a} as  c^{x\log_c a}\geq2
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  3. #3
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    I dont understand how this answers the question. I remember the idea was to show that Z < 2*XlogA and Z < 2*YlogB and so Z<2*max(XlogA, YlogB).

    chiph588@, I dont see how XlogA > YlogB. Nor do I understand how A^X + B^Y is equal to A^X+ B^Y=c^X^l^o^gc^A  + c^Y^l^o^gc^A
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chauhan View Post
    I dont understand how this answers the question. I remember the idea was to show that Z < 2*XlogA and Z < 2*YlogB and so Z<2*max(XlogA, YlogB).

    chiph588@, I dont see how XlogA > YlogB. Nor do I understand how A^X + B^Y is equal to A^X+ B^Y=c^X^l^o^gc^A  + c^Y^l^o^gc^A
    I just assumed one was bigger than the other.

    Also  a^x=c^{\log_c a^x} = c^{x\log_c a} .
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