Does $^{\frac12}{a}=$ $\sqrt{a}_s$?

Question

Just to be sure, does $^{\frac12}{a}=$ $\sqrt{a}_s$? I only ask because, although the Wikipedia page on tetration and other sources explain that the super-root is one of the inverse operations of tetration, I have never seen the equation "$^{\frac12}{a}=$ $\sqrt{a}_s$" shown as another way to express the square super-root or any other super-root.

I bring this up realizing that the properties of exponentiation and its inverses do not completely fall in line with tetration and its inverses. For example, generally, $^{bc}{a}≠$ $^{b}({^c{a})}$. Contrastingly, if I'm not mistaken, $^{\operatorname{slog}_ab}a=b$ and ${\operatorname{slog}_a(^{b}a)}=b$, which seems to mimic $a^{\log_ab}=b$ and $\log_a(a^b)=b$.

I understand this may seem like the most trivial of questions, but I never like assuming things without knowing for sure, especially with hyperoperations above exponentiation.

Answer 1

I'm assuming that by the super-root you mean that when $y=x^x$, $x$ is the super-root of y.

Then $^{\frac12}{a}= \sqrt{a}_s$ gives ${(^{\frac12}{a})}^{(^{\frac12}{a})}=a$

Squaring gives $a^a = 2^a a^2$

Wolfram Alpha gives $a=4$ and $a \approx 0.692646...$ .

So $^{\frac12}{a}= \sqrt{a}_s$ can't be an identity.

More simply, let $a=1$, then since $^k1=1$ for $k\in\mathbb{N}$, $\sqrt{1}_s=1$, while $^{\frac12}{a}$ goes to $^\frac12$.

Once again $^{\frac12}{a}= \sqrt{a}_s$ can't be an identity.