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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- Jul 12th 2010, 09:04 PMchauhanWhy 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)?? - Jul 12th 2010, 09:14 PMchiph588@
- Jul 12th 2010, 09:48 PMchauhan
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 $\displaystyle A^X+ B^Y=c^X^l^o^gc^A + c^Y^l^o^gc^A$ - Jul 12th 2010, 09:55 PMchiph588@