May 30th, 2014, 12:04 AM | #1 |
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8 | solve |
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 |
Newbie Joined: May 2014 From: China Posts: 4 Thanks: 4 |
here is my answer. |
May 30th, 2014, 01:29 AM | #4 |
Senior Member Joined: Aug 2011 Posts: 334 Thanks: 8 | |
May 30th, 2014, 01:46 AM | #5 | |
Newbie Joined: May 2014 From: China Posts: 4 Thanks: 4 | i think you may ignore that as follows. Quote: | |
May 30th, 2014, 01:50 AM | #6 |
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 | |
May 30th, 2014, 07:05 AM | #7 |
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 | The answer is right. I'll eat my words. $\displaystyle 2012\cdot { 2012 }^{ 2012 }={ 2012 }^{ 2013 }$ See these two links: *(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. |