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.
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.
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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Hello, dwsmith!
$\displaystyle \text{Find }x \text{ if: }\:x^{x^{x^{\hdots}}}\:=\:2$
$\displaystyle \text{We are given: }\;x^{x^{x^{\hdots}}} \:=\:2$
$\displaystyle \text{Then: }\;x^{\left(x^{x^{\hdots}}\right)\;\leftarrow\;\te xt{This is 2}} $
$\displaystyle \text{The equation becomes: }\:x^2 \:=\:2$
$\displaystyle \text{Therefore: }\:x \:=\:\sqrt{2}$