Why this inequation?

• Jul 12th 2010, 09:04 PM
chauhan
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)??
• Jul 12th 2010, 09:14 PM
chiph588@
Quote:

Originally Posted by chauhan
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 $c^{x\log_c a}\geq2$
• Jul 12th 2010, 09:48 PM
chauhan
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 $A^X+ B^Y=c^X^l^o^gc^A + c^Y^l^o^gc^A$
• Jul 12th 2010, 09:55 PM
chiph588@
Quote:

Originally Posted by chauhan
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 $a^x=c^{\log_c a^x} = c^{x\log_c a}$.