$\sqrt{\text{Square root}}$

$\huge(\sqrt{2})^{\sqrt{2} \times\sqrt{2}}=(\sqrt{2})^{2}=\boxed{2}$

Let x= ((squroot 2)^squroot 2)^squroot 2. Now take logarithms. So Log x = log (((squroot 2)^squroot 2)^squroot 2). Hence Log x = (squroot 2) x log ((squroot 2)^squroot 2). (This is a property of logarithms!). And so hence Log x = (squroot 2) x (squroot 2 x log (squroot 2). Hence Log x = 2 x log (squroot 2). Now do the reverse process. Hence Log x = log ((squroot 2)^2). Hence log x = log 2. Hence x = 2. Regards, David

(2^1/2)^(2^1/2)^(2^1/2)=(2^1/2)^2 =2^({1/2}*2) =2^1 =2

(2^1/2)^(2^1/2)^(2^1/2)=(2^1/2)^2 =2^({1/2}*2) =2^1 =2

Put expression ((√2)^√2)√2=y apply (ln) function for two sides so √2ln(√2)^√2=ln(y) =√2 √2ln(2)^1/2= 1/2 (2)*ln(2)=ln(y) so y=2######

When evaluating towering exponents, we use the following rule:

a^b^c = a^bc

Therefore, (sqrt(2))^sqrt(2)^sqrt(2) = sqrt(2)^sqrt(2) • sqrt(2) = sqrt(2)^2 = 2