Find $x$ in this algebra

Algebra Level 1

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

\sqrt{2}

\infty

1 solution

Phạm Hoàng
May 25, 2018

First call $x^{x^{x^{...}}}=p$ . $p=2$ .Next $x^{p}=2$ .So $x=\sqrt{2}$ .

If $p=4$ , then $x=\sqrt2$ as well! So what is the value of $x^{x^{x^{...}}}$ if $x=\sqrt2$ ? 2 or 4?

Pi Han Goh - 3 years ago

Great question!If you log on your calculator,it's actually approach to 2.But the question said you find $x$ ,not the value.

Phạm Hoàng - 3 years ago