November 10th, 2016, 05:55 PM  #11 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,659 Thanks: 652 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, 06:03 PM  #12 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond 
I agree and echo Dan's request.

November 11th, 2016, 11:57 AM  #13  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
 
November 11th, 2016, 12:29 PM  #14  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,091 Thanks: 2360 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, 01:02 PM  #15 
Global Moderator Joined: Dec 2006 Posts: 18,245 Thanks: 1439 
The post and cartoon related to finding integer solutions, but that wasn't mentioned.

December 26th, 2016, 12:27 PM  #16 
Senior Member Joined: Dec 2016 From: Italy Posts: 252 Thanks: 1  Solution of equation
.
Last edited by anyelo; December 26th, 2016 at 12:30 PM. 
December 26th, 2016, 01:43 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 01:46 PM. 

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