# Challenge question

• Apr 19th 2010, 08:06 PM
dwsmith
Challenge question
Find x if $x^{x^{x^{...}}}=2$.

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 PM
Bruno J.
Well clearly $x^2 = x^{(x^{x^{x^{\dots}}})}=2$, so $x=\sqrt 2$.
• Apr 19th 2010, 08:23 PM
dwsmith
Quote:

Originally Posted by Bruno J.
Well clearly $x^2 = x^{(x^{x^{x^{\dots}}})}=2$, so $x=\sqrt 2$.

Correct, I did it like this.

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

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

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

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

$aln(x)=ln(2)$

$2ln(x)=ln(2)$

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

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

$x=\sqrt{2}$
• Apr 26th 2010, 07:18 AM
kandyfloss
which log rules did you use here?
• Apr 26th 2010, 04:13 PM
dwsmith
Quote:

Originally Posted by kandyfloss
which log rules did you use here?

Exponential
• Apr 30th 2010, 07:45 PM
hollywood
Quote:

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.

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 PM
Soroban
Hello, dwsmith!

Quote:

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

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

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

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

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

• May 18th 2010, 01:18 AM
chirkowatson
Quote:

Originally Posted by dwsmith
Correct, I did it like this.

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

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

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

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

$aln(x)=ln(2)$

$2ln(x)=ln(2)$

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

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

$x=\sqrt{2}$

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.
• May 21st 2010, 10:08 AM
Ted
Quote:

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

Anyone could explain it for me !!
why this is 2 !!
• May 21st 2010, 10:20 AM
Quote:

Originally Posted by Ted
Anyone could explain it for me !!
why this is 2 !!

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

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

Since it's an infinite power tower of x's you can do that...
• Feb 8th 2011, 10:47 AM
dwsmith
Quote:

Originally Posted by Ted
Anyone could explain it for me !!
why this is 2 !!

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

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