Tower of exponents solution approach unique to exponents?

[DocZaius]

I saw a YouTube video presenting what seemed like a clever solution to $x^{x^{x^{.^{.}}}} = 2$ (which is to say: an infinite tower of exponents of x = 2). He said to consider just the exponents and ignore the base and realize that those exponents themselves become a restatement of the whole left side of the equation. Equate them to 2 and you get $x^2=2$ and thus $x= \sqrt{2}$.

I thought that was clever and tried to see if this could work for other operations like multiplication ($x\cdot x\cdot x \cdot\ ...=2$), division ($\frac{x}{\frac{x}{\frac{x}{...}}}=2$), addition ($x+x+x+...=2$) and subtraction($x-x-x-...=2$), but realized it didn't. Any idea what's special about the exponent operation that allows for this clever "substitution of infinite terms" type of solution?

 

[mfb]

The exponent tower can have a limit, the other processes do not (apart from boring special cases).

That should work as well:
$$x\sqrt{x\sqrt{x\sqrt \dots}}$$

 

[Samy_A]

I saw a YouTube video presenting what seemed like a clever solution to $x^{x^{x^{.^{.}}}} = 2$ (which is to say: an infinite tower of exponents of x = 2). He said to consider just the exponents and ignore the base and realize that those exponents themselves become a restatement of the whole left side of the equation. Equate them to 2 and you get $x^2=2$ and thus $x= \sqrt{2}$.

I thought that was clever and tried to see if this could work for other operations like multiplication ($x\cdot x\cdot x \cdot\ ...=2$), division ($\frac{x}{\frac{x}{\frac{x}{...}}}=2$), addition ($x+x+x+...=2$) and subtraction($x-x-x-...=2$), but realized it didn't. Any idea what's special about the exponent operation that allows for this clever "substitution of infinite terms" type of solution?

Even for the infinite tower of exponents the trick will only work if $e^{-e}<x<e^{1/e}$. As $e^{1/e}=1.444...$ and $\sqrt 2=1.414...$, it works for $\sqrt 2$.
https://thatsmaths.files.wordpress.com/2013/01/powertowerlambert.pdf