# Math Help - Challenge question

1. ## 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.

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

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

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

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

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

$aln(x)=ln(2)$

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

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

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

$x=\sqrt{2}$

4. which log rules did you use here?

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

6. 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".

7. Hello, dwsmith!

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

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

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

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

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

8. 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.

9. 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 !!

10. 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...

11. 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.