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 May 30th, 2014, 12:04 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 solve Thanks from raul21
 May 30th, 2014, 01:08 AM #2 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology e^ln{x^x} = e^(ln(2012^2012^2013)) x lnx = ln(2012^2012^2013) (because e^x is an injection) x lnx = 2013 ln (2012^2012) x lnx = 2013 mult. 2012 ln 2012 x lnx = 4050156 ln 2012 that's all i could menage for now...am i even going in the right direction?
May 30th, 2014, 01:26 AM   #3
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May 30th, 2014, 01:29 AM   #4
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Wat is the interest of such computation ?

http://http://www.wolframalpha.com/i...82012%29+for+x
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May 30th, 2014, 01:46 AM   #5
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i think you may ignore that as follows.

Quote:
 Originally Posted by JJacquelin Wat is the interest of such computation ? http://http://www.wolframalpha.com/i...82012%29+for+x
i think you may ignore that as follows.
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May 30th, 2014, 01:50 AM   #6
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Quote:
 Originally Posted by ldxylizhi here is my answer.
Congrats. i just forgot to apply ln once again.

May 30th, 2014, 07:05 AM   #7
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Quote:
 Originally Posted by ldxylizhi here is my answer.
The answer is right. I'll eat my words.

$\displaystyle 2012\cdot { 2012 }^{ 2012 }={ 2012 }^{ 2013 }$

*(2012^2012)^(2012^2012) - Wolfram|Alpha

*2012^(2012^2013) - Wolfram|Alpha

Last edited by perfect_world; May 30th, 2014 at 07:38 AM.

 May 30th, 2014, 07:32 AM #8 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 actually, it's correct. lol. Got caught out by the simplicity of it. Doh! Last edited by perfect_world; May 30th, 2014 at 07:47 AM.

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