I was looking at a problem that deals with tetration. Here's the problem: $$\begin{split}x^{x^{x^{\ddots}}}=3 \\ x^3=3 \Rightarrow x= \sqrt[3]{3} \end{split}$$ However, I saw that the solution is $-\frac{3}{\ln(3)}W_{-1}\left(-\frac{3}{\ln(3)}\right)\approx 2.478$ and watched some videos which explain something like the convergence "test" of $e^{-e}$ and $e^e$ and more ones some Lambert $W_{-1}$ function but I don't undertand this one.
I try to solve that one by this logic: $$x^{x^{x^{\ddots}}}=n$$ So $$x^n=n\text{, which implies, $\forall n \in \mathbb{R}$,}\, x=\sqrt[n]{n}\text{.}$$
What's wrong with this logic?
The answer is simpler and no knowledge about the Lambert function is necessary.
The infinite exponential tower $$x^{x^{x^\cdots}}$$ is defined as the limit of the sequence $$x,x^x,x^{x^x},x^{x^{x^x}},\cdots$$ when this sequence converge. This happens to be for $x \in [e^{-e},e^{1/e}]$. So those are the only real values of $x$ for which the expression $$x^{x^{x^\cdots}}$$ makes sense. That means that $$f(x)=x^{x^{x^\cdots}}$$ defines a function $f:[e^{-e},e^{1/e}] \to \Bbb R$. The image of that function turn out to be $[1/e,e]$.
Since $3\not\in [1/e,e]$, that simply means that $$x^{x^{x^\cdots}} = 3$$ has no solution.
But your reasoning works if you choose a number in the image of the function. For any $y\in [1/e,e]$ the solution of $$x^{x^{x^\cdots}} = y$$ is in fact $x=\sqrt[y]y$.