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 $a^x=c^{\log_c a^x} = c^{x\log_c a}$.
The content structure of the text passage involves a mathematical question about an inequality involving logarithms and exponents, followed by a response that attempts to provide a proof or explanation for the inequality.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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 $a^x=c^{\log_c a^x} = c^{x\log_c a}$.
user
The content structure of the text passage involves a mathematical question about an inequality involving logarithms and exponents, followed by a response that attempts to provide a proof or explanation for the inequality.

TYPE: SINGLE_QUESTION_WITH_ANSWER