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.

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- Apr 19th 2010, 08:06 PMdwsmithChallenge 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. - Apr 19th 2010, 08:22 PMBruno J.
Well clearly $\displaystyle x^2 = x^{(x^{x^{x^{\dots}}})}=2$, so $\displaystyle x=\sqrt 2$.

- Apr 19th 2010, 08:23 PMdwsmith
Correct, I did it like this.

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

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

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

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

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

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

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

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

$\displaystyle x=\sqrt{2}$ - Apr 26th 2010, 07:18 AMkandyfloss
which log rules did you use here?

- Apr 26th 2010, 04:13 PMdwsmith
- Apr 30th 2010, 07:45 PMhollywood
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". - Apr 30th 2010, 08:29 PMSoroban
Hello, dwsmith!

Quote:

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

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

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

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

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

- May 18th 2010, 01:18 AMchirkowatson
- May 21st 2010, 10:08 AMTed
- May 21st 2010, 10:20 AMDeadstar
- Feb 8th 2011, 10:47 AMdwsmith