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November 10th, 2016, 05:55 PM   #11
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Can I ask you guys to stop arguing on the open forum? It's made a mess of several threads now.

-Dan
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November 10th, 2016, 06:03 PM   #12
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I agree and echo Dan's request.
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November 11th, 2016, 11:57 AM   #13
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Quote:
Originally Posted by v8archie View Post
Physician, heal thyself.

I have zero need to heal myself. Again, you don't know what you're talking about.

I still haven't looked at your cartoon because I'm not interested in it.
You're *clearly* lying, else you wouldn't have made the comment about "attempt at humor."

I'm under no obligation to follow and read any links posted by anybody here.

No one stated you were. You're bringing up a cure for which there was no illness.


If you have any interesting comment to make on the topic, go ahead an [sic] post it.

You never should have made yours after my first post in the first place.

In the absence of the above, I believe this conversation is done.
You were absent from the truth. That's why I posted this.
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November 11th, 2016, 12:29 PM   #14
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Originally Posted by Math Message Board tutor View Post
You're *clearly* lying, else you wouldn't have made the comment about "attempt at humor."
No, I just surmised that your posting of a link to a cartoon would be an attempt at humour.

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Originally Posted by Math Message Board tutor View Post
You never should have made [your post] after my first post in the first place.
I have every right to post mathematical comments about mathematical posts on this site. I repeat, if you have any disagreement about the mathematical content of my post, we can discuss it here. Otherwise, let's call a halt to the pointless bickering.

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.
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November 11th, 2016, 01:02 PM   #15
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The post and cartoon related to finding integer solutions, but that wasn't mentioned.
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December 26th, 2016, 12:27 PM   #16
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Solution of equation

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Last edited by anyelo; December 26th, 2016 at 12:30 PM.
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December 26th, 2016, 01:43 PM   #17
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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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