November 10th, 2016, 04:55 PM  #11 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,686 Thanks: 666 Math Focus: Wibbly wobbly timeywimey stuff. 
Can I ask you guys to stop arguing on the open forum? It's made a mess of several threads now. Dan 
November 10th, 2016, 05:03 PM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,757 Thanks: 1008 Math Focus: Elementary mathematics and beyond 
I agree and echo Dan's request.

November 11th, 2016, 10:57 AM  #13  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
 
November 11th, 2016, 11:29 AM  #14  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,214 Thanks: 2410 Math Focus: Mainly analysis and algebra  Quote:
Quote:
If you are really serious about keeping this argument going, open up your account to receive private messages, and we can go for it without bothering other people.  
November 11th, 2016, 12:02 PM  #15 
Global Moderator Joined: Dec 2006 Posts: 18,686 Thanks: 1522 
The post and cartoon related to finding integer solutions, but that wasn't mentioned.

December 26th, 2016, 11:27 AM  #16 
Senior Member Joined: Dec 2016 From: Italy Posts: 252 Thanks: 1  Solution of equation
.
Last edited by anyelo; December 26th, 2016 at 11:30 AM. 
December 26th, 2016, 12:43 PM  #17 
Senior Member Joined: Dec 2016 From: Italy Posts: 252 Thanks: 1 
Let $\displaystyle \:\:\: t =\frac{\log_3 x  2}{2} \:\:\:\:\:\:\:\: \Longrightarrow \:\:\:\:\:\:\:\: x =3^{2t+2} $ $\displaystyle x^{\log_3 2} =\sqrt x+1 $ $\displaystyle \left({3^{2t+2}}\right)^{\log_3 2} =\sqrt {3^{2t+2}}+1 $ $\displaystyle 3^{\log_3 2\cdot(2t+2)} =3^{t+1}+1 $ $\displaystyle \left({3^{\log_3 2}}\right)^{2t+2} =3^{t+1}+1 $ $\displaystyle 2^{2t+2} =3^{t+1}+1 $ $\displaystyle 4^{t+1} =3^{t+1}+1 $ $\displaystyle t > 0 \:\:\:\:\: \Rightarrow \:\:\:\:\: 4^t > 3^t \:\:\:\land \:\:\: 4^t > 1 \:\:\:\:\: \Rightarrow \:\:\:\:\: 3\cdot4^t > 3\cdot3^t \:\:\:\:\land \:\:\:\: 4^t > 1 \:\:\:\:\: \Rightarrow \:\:\:\:\: 3\cdot4^t +4^t> 3\cdot3^t +1\:\:\:\:\: \Rightarrow \:\:\:\:\: 4^{t+1}> 3^{t+1} +1 $ $\displaystyle t < 0 \:\:\:\:\: \Rightarrow \:\:\:\:\: 4^t < 3^t \:\:\:\land \:\:\: 4^t < 1 \:\:\:\:\: \Rightarrow \:\:\:\:\: 3\cdot4^t < 3\cdot3^t \:\:\:\:\land \:\:\:\: 4^t < 1 \:\:\:\:\: \Rightarrow \:\:\:\:\: 3\cdot4^t +4^t < 3\cdot3^t +1\:\:\:\:\: \Rightarrow \:\:\:\:\: 4^{t+1} < 3^{t+1} +1 $ So $\displaystyle \:\: t = 0 \:\: $ is the unique solution of the equation $\displaystyle \:\: 4^{t+1} =3^{t+1}+1 $ It means that $\displaystyle \:\: x = 3^{2\cdot0+2} = 9\:\: $ is the unique solution of the equation $\displaystyle \:\: x^{\log_3 2} =\sqrt x+1 $ Last edited by anyelo; December 26th, 2016 at 12:46 PM. 

Tags 
hard, logarithm, problem, super 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Hard logarithm question  IBstudent  Algebra  5  May 31st, 2017 02:37 AM 
Super easy probability problem  CyberRocket  Probability and Statistics  1  July 18th, 2016 09:41 AM 
Logarithm Problem  Kiphyl  PreCalculus  7  August 1st, 2014 09:54 PM 
super simple algebra problem.  questioner1  Algebra  5  July 2nd, 2012 09:59 PM 
super hard 16 year old CIRCLE THEOREMS questions please help  peterle1  Algebra  2  February 11th, 2010 10:46 AM 