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April 15th, 2012, 05:14 AM  #1 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Another logarithmic diophantine problem !!!
How to prove that x^y = y^x (x is not = y) has no solution if y > 2 ????

May 25th, 2012, 04:21 AM  #2 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Another logarithmic diophantine problem !!!
I found a method for reducing it into normal diophantine form: x^y=y^x, assume x,y is very large (x,y integer) then y/ln(y) = x/ln(x) implies p(y)=p(x) where p is the prime counting function. is it possible to solve from here!? 
May 25th, 2012, 05:19 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1981 
As it's not true, there's no way to prove it.

May 25th, 2012, 05:56 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond  Re: Another logarithmic diophantine problem !!!
Can you provide a counterexample?

May 25th, 2012, 06:42 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1981 
x = 2, y = 4 (there are infinitely many others if x and y needn't be integers).

May 25th, 2012, 11:17 PM  #6  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Another logarithmic diophantine problem !!! Quote:
 
May 26th, 2012, 07:49 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,310 Thanks: 1981 
Example of noninteger solution: x = 2^(2/3), y = 2^(8/3).

May 26th, 2012, 10:19 AM  #8  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Another logarithmic diophantine problem !!! Quote:
now I'm asking you again, how to prove the above statement without the use of calculus?  
September 22nd, 2012, 08:02 AM  #10  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Another logarithmic diophantine problem !!! Quote:
Perhaps this would be appropriate if posted to the NT section. I posted this after a few months of joining so I wasn't familiar enough with this forum. If moderators agree, then I could create another topic in the NT section asking about a Diophantine method to solve x^y = y^x where x,y are integers. Then mods can post a link to that topic and lock this topic. I don't want it to be moved since it's already been months before I posted it and due to the discussion here, people can easily get confused about what I am exactly asking. Please inform me if any of you moderators agree with this. Balarka .  

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