Challenge question

Find x if $\displaystyle x^{x^{x^{...}}}=2$.

I sent the answer to about 5 people since I have no idea how to figure out who all the moderators are.



Moderator edit: Approved challenge question.

Well clearly $\displaystyle x^2 = x^{(x^{x^{x^{\dots}}})}=2$, so $\displaystyle x=\sqrt 2$.

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Originally Posted by Bruno J.

Well clearly $\displaystyle x^2 = x^{(x^{x^{x^{\dots}}})}=2$, so $\displaystyle x=\sqrt 2$.

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Correct, I did it like this.

Find x if $\displaystyle x^{x^{x^{...}}}=2$.

$\displaystyle a=x^{x^{...}}=2$

$\displaystyle ln(x^{x^{...}})=ln(2)$

$\displaystyle x^{x^{...}}ln(x)=ln(2)$

$\displaystyle aln(x)=ln(2)$

$\displaystyle 2ln(x)=ln(2)$

$\displaystyle ln(x)=\frac{1}{2}ln(2)$

$\displaystyle ln(x)=ln(\sqrt{2})$

$\displaystyle x=\sqrt{2}$

which log rules did you use here?

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Originally Posted by kandyfloss

which log rules did you use here?

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Exponential

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Originally Posted by dwsmith

I sent the answer to about 5 people since I have no idea how to figure out who all the moderators are.

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The list of moderators is here:
http://www.mathhelpforum.com/math-help/showgroups.php
Don't be tricked by the "Moderators" heading - they are down at the bottom under "MHF Moderators".

Hello, dwsmith!

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$\displaystyle \text{Find }x \text{ if: }\:x^{x^{x^{\hdots}}}\:=\:2$

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$\displaystyle \text{We are given: }\;x^{x^{x^{\hdots}}} \:=\:2$

$\displaystyle \text{Then: }\;x^{\left(x^{x^{\hdots}}\right)\;\leftarrow\;\te xt{This is 2}} $

$\displaystyle \text{The equation becomes: }\:x^2 \:=\:2$

$\displaystyle \text{Therefore: }\:x \:=\:\sqrt{2}$

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Originally Posted by dwsmith

Correct, I did it like this.

Find x if $\displaystyle x^{x^{x^{...}}}=2$.

$\displaystyle a=x^{x^{...}}=2$

$\displaystyle ln(x^{x^{...}})=ln(2)$

$\displaystyle x^{x^{...}}ln(x)=ln(2)$

$\displaystyle aln(x)=ln(2)$

$\displaystyle 2ln(x)=ln(2)$

$\displaystyle ln(x)=\frac{1}{2}ln(2)$

$\displaystyle ln(x)=ln(\sqrt{2})$

$\displaystyle x=\sqrt{2}$

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I am completely supporting to his method as exponential method is perfect. I suggest you to this same method is not twisted and very easy to remember.

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Originally Posted by Soroban

$\displaystyle \text{Then: }\;x^{\left(x^{x^{\hdots}}\right)\;\leftarrow\;\te xt{This is 2}} $

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Anyone could explain it for me !!
why this is 2 !!

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Originally Posted by Ted

Anyone could explain it for me !!
why this is 2 !!

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$\displaystyle x^{x^{x^{\hdots}}} = 2$

so $\displaystyle x^{\big{(}x^{x^{x^{\hdots}}}\big{)}} = x^2$

Since it's an infinite power tower of x's you can do that...

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Originally Posted by Ted

Anyone could explain it for me !!
why this is 2 !!

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$\displaystyle x^{x^{x^{...}}}=2$

If we isolate an x, we still have an infinite x to the x which is stated as being equal to 2.