Is this a valid proof for ${x^{x^{x^{x^{x^{\dots}}}}}} = y$?

Question

So I got this challenge from my teacher.

Solve ${x^{x^{x^{x^{x^{\dots}}}}}} = y$ (eq. 1) for $x$.

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My attempt:

As $x^{y^z}$ per definition equals $x^{y \cdot z}$, then $x^y = y$ from (eq. 1). Thus, $x = \sqrt{y}$.

Is this a valid proof?

Answer 1

It's not a correct proof, no.

If $${x^{x^{x^{x^{x^{\dots}}}}}} = y$$ then we can say that $x$ to the power of each side is the same:

$$x^{\left({x^{x^{x^{x^{x^{\dots}}}}}}\right)} = x^y$$

but then the left hand side is what exactly what we started with (provided the limit exists) so we can equate the right hand sides of each of these equations:

$$y=x^y\implies x=y^{1/y}$$

Answer 2

Really? $x^{y^z}$ equals $x^{y\cdot z}$? So you are saying that $$2=2^1=2^{1^2} = 2^{1\cdot 2} = 2^2 = 4?$$

Answer 3

Once you prove that over the interval $\left[e^{-e},e^{\frac{1}{e}}\right]$: $$ f(x)=x^{x^{x^{x^{\ldots}}}} = \frac{W(-\log x)}{-\log x}\tag{1}$$ where $W$ is the Lambert W-function, it follows that: $$ f^{-1}(x) = x^{\frac{1}{x}}.\tag{2} $$ On the other hand, $$ x = x^{f(x)} \tag{3} $$ implies: $$ x = f^{-1}(x)^x \tag{4}$$ hence $(2)$ is quite trivial.