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September 18th, 2018, 01:55 AM  #1 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  I need help please
How to find x in this equation: x^x^x=27 I have tried different methods to solve that with no success. I tried to take logarithm from both sides but it ended up with mess. I put x^x=y and tried to solve x^y=3^3 but nothing further after that. 
September 18th, 2018, 03:16 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics 
Apply Newton's method to the function $f(x) = x^{x^x}  27$.

September 18th, 2018, 04:41 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Where did you get this problem from?

September 18th, 2018, 05:55 AM  #4 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  
September 18th, 2018, 07:29 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Does your daughter know where it came from or did she make it up?

September 18th, 2018, 07:34 AM  #6 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  
September 18th, 2018, 08:10 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
Could the question have been x^x = 27 originally, but got changed in the course of being passed on from person to person?

September 18th, 2018, 09:06 AM  #8 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  
September 18th, 2018, 05:50 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,966 Thanks: 2216 
In that case, there doesn't seem to be a closed expression that gives x explicitly.

September 26th, 2018, 04:49 PM  #10 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 661 Thanks: 87 
This isn't an exact answer, but my graphing calculator says x is about 2.1046.
