Nested Rooted 2's

Let $(\sqrt2)^{{(\sqrt2)^{\ldots}}} = x$

$x = (\sqrt2)^x=2^{\frac{x}{2}}$

$\Rightarrow x=2^{\frac{x}{2}}$

Squaring Both Sides,

$x^2=2^x$

Comparing both sides, $\boxed{x=2}$

misalkan V2^V2^V2^... =x (V2)^x=x 2^x = x^2 so,x=2

Let $x=(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{\ldots}}}$ . Then $x=(2)^{\left(\frac{\sqrt{2}}{2}\right)^{(\sqrt{2})^{\ldots}}}$ $x=(2)^{\left(\frac{1}{\sqrt{2}}\right)^{(\sqrt{2})^{\ldots}}}$ $x=(2)^{\left(\frac{1^{\left(\sqrt{2}\right)^{\ldots}}}{\sqrt{2}^{\left(\sqrt{2}\right)^{\ldots}}}\right)}$ $x=(2)^{\left(\frac{1}{x}\right)}$ Take the log of both sides gives $\ln{x}=\ln{(2)^{\left(\frac{1}{x}\right)}}$ $\ln{x}={\left(\frac{1}{x}\right)}\ln{(2)}$ $x\ln{x}=\ln{(2)}$ $\ln{x^{x}}=\ln{(2)}$ Raise both sides as the powers of common base $e$ to get $x^{x}=(2)$ But note that due to the infinite nature of the exponents, $x^{x}=(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{(\sqrt{2})^{\ldots}}}}}}=x$ So by the transitive property, $x^{x}=x=2$

Let \sqrt{2}^{x}= 2^{x/2} Log 2{x}= x/2 log 2{2} 2 log 2{x}=xlog 2{2} Comparing both the side... x=2