text: 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 $a^x=c^{\log_c a^x} = c^{x\log_c a}$.
explanation: The document contains a math question and a subsequent discussion that attempts to answer the question. The question is about understanding an inequality involving logarithms and exponents. The discussion includes a proposed solution and a follow-up clarification request from the original poster.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: 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 $a^x=c^{\log_c a^x} = c^{x\log_c a}$.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a subsequent discussion that attempts to answer the question. The question is about understanding an inequality involving logarithms and exponents. The discussion includes a proposed solution and a follow-up clarification request from the original poster.
  
  Label: question-and-answer