What is the inverse function of $x^{x^x}....$ x times?

Question

Let $M(x) = x^{x^x}...$ x times. How would I extend this function to take on decimals and negatives inputs. I feel like this could be done in a way that makes sense similar to how $x^x$ is easily extended (with decimals like 0.5 being roots and so on). Also, what is $M^{-1}(x)$? Thanks in advance.

Answer 1

The inverse operations of tetrations are noted as the super-logarithm and the super-root. $$\large{M=\;^nx=x^{x^{.^{.^x}}}}$$ Super-Logarithm

For all real numbers $n$ and $x\gt 1$, we have $$\large{\operatorname{slog}_x\left(M\right)=\operatorname{slog}_x\left(^nx\right)=n}$$

Super-Root

For each integer $n\gt 2$ and $x\geq 1$, we have $$\large{\sqrt[n]{M}_s=\sqrt[n]{^nx}_s=x}$$

Square Super-Root

For $n=2$ and $x\geq 1$, we have $$\large{\sqrt{M}_s=\sqrt{^2x}_s=x}$$

Hope this helps.