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November 8th, 2016, 11:36 AM   #1
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Lightbulb A super hard logarithm problem

Hey, it's my first time visiting in a math forum but I seriously need help.
Since there are no active forums in hebrew (I'm from israel) I went looking for english forums, like this one.
So my math teacher gave it to us today saying he will bump up the grade of anyone solving it, showing they way of solving, obviously.
Anyway, this is the problem:

Anyone solving or at least trying to solve it, be blessed! thank you.
*explain how you did it. the final result is 9, as he told us, but the way is what matters.
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November 8th, 2016, 02:20 PM   #2
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Quote:
Originally Posted by DonTry View Post
Hey, it's my first time visiting in a math forum but I seriously need help.
Since there are no active forums in hebrew (I'm from israel) I went looking for english forums, like this one.
So my math teacher gave it to us today saying he will bump up the grade of anyone solving it, showing they way of solving, obviously.
Anyway, this is the problem:

Anyone solving or at least trying to solve it, be blessed! thank you.
*explain how you did it. the final result is 9, as he told us, but the way is what matters.
Graphing seems to give x = 9. There is no method to solve this save by approximation.

-Dan
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November 8th, 2016, 04:58 PM   #3
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After some manipulation and using some of the rules of logarithms one may arrive at

$$\left(\frac23\right)^y=\frac{\sqrt x+1}{x}$$

which has an integer solution at $x=9$.
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November 8th, 2016, 05:16 PM   #4
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Let x = 9^t, so that the equation becomes 4^t = 3^t + 1.
By inspection, t = 1 is a solution, and so x = 9.
Show that no other solution exists.
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November 9th, 2016, 09:17 AM   #5
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Let x = 9^t, so that the equation becomes 4^t = 3^t + 1.
By inspection, t = 1 is a solution, and so x = 9.
Show that no other solution exists.
$\displaystyle 4^t - 1 \ = \ 3^t$

$\displaystyle (2^t - 1)(2^t + 1) \ = \ 3^t$

For possible integer solutions,

$\displaystyle (2^t - 1)(2^t + 1) \ = \ 3\cdot3^{t - 1}$


$\displaystyle 2^t - 1 \ = \ 3 $
$\displaystyle 2^t + 1 \ = \ 3^{t - 1}$
-----------------


or it's the case


$\displaystyle 2^t - 1 \ = \ 3^{t - 1}$
$\displaystyle 2^t + 1 \ = \ 3 $
-----------------


One of the above scenarios isn't possible, while the other works.
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November 9th, 2016, 09:44 AM   #6
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Originally Posted by Math Message Board tutor View Post
$\displaystyle (2^t - 1)(2^t + 1) \ = \ 3\cdot3^{t - 1}$
What follows the quoted line is only true for one particular choice of $t$. It is not in general true that, given two factorizations of an expression into two terms that the factors must be equal.
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November 9th, 2016, 02:50 PM   #7
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What follows the quoted line is only true for one particular choice of $t$.
It is not in general true that, given two factorizations of an expression into two terms that the factors must be equal.
my reply:

Mymathforum.com - - - 11/09/2016 by newuser789 : stripcreator
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November 9th, 2016, 03:14 PM   #8
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I'm not interested in your attempts at humour. If you have something to say to me you can say it here.
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November 10th, 2016, 05:05 PM   #9
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I'm not interested in your attempts at humour. If you have something to say to me you can say it here.
I put the reply in the first panel to show what a presumptuous poster you are! If you're too darned lazy to R E A D it,
then don't act as if you didn't read it and then make a comment on it to show that you in fact did read it!


Maybe next time you will reconsider making a post to someone that is not presumptuous. Hmm?

Last edited by Math Message Board tutor; November 10th, 2016 at 05:08 PM.
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November 10th, 2016, 05:37 PM   #10
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Originally Posted by Math Message Board tutor View Post
Maybe next time you will reconsider making a post to someone that is not presumptuous.
Physician, heal thyself.

I still haven't looked at your cartoon because I'm not interested in it. I'm under no obligation to follow and read any links posted by anybody here.

If you take issue with the accuracy of my post #6, go ahead and explain what you think is wrong with it.

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

In the absence of the above, I believe this conversation is done.
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