Fun 2

Algebra Level 2

What is the above expression equivalent to ?

2 solutions

Vinayak Srivastava
May 14, 2020

Let ${{{\sqrt2}^{\sqrt2}}^{\sqrt2}}^{\cdots}=x$

So, ${\sqrt2}^x=x$ .

Hence, ${\sqrt[2]2}=\sqrt[x]{x}$

Clearly, $\boxed{x=2}$

Fox To-ong
Feb 9, 2015

that equation will become (2^0.50)^0.50 = 2^1 = 2

This solution is wrong as ${\sqrt2}^{\sqrt2}\neq{2}$

Vinayak Srivastava - 1 year, 1 month ago