Explain $x^{x^{x^{{\cdots}}}} = \,\,3$ $$x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} =\quad 2$$
This equation has the answer $\sqrt{2}$ by taking $\log$ to both side. This answer is correct because I'd proven it by replacing $x$ with $\sqrt{2}$ then computed the expression repeatedly and the answer is close to $2$.
But now consider,$$x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} =\quad 3$$
I tried using the same solving method and it gave me that $\sqrt[3]{3}$ is the answer but when I tried proving the answer by replacing $x$ with $\sqrt[3]{3}$ and computed it repeatedly, the answer is closer to $2.478052680288297$ and not $3$
I don't know why this happens, could someone explain the answer to $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 3$ ?
I don't think it's $\sqrt[3]{3}$
The first equation problem came from brilliant.org

There is one more observation which should be useful for you.
Begin again with the problem of iterated exponentiation to base $b=\sqrt2$. If you start with some value smaller than $x_0 \lt 4$ you arrive at $x_\infty=t_0=2$:                          
$ \quad \qquad \displaystyle  x_0=1.1 \\
 \quad \qquad   x_k = b ^{x_{k-1}} \qquad \qquad \text{ where } b=\sqrt 2\\
 \quad \qquad   \lim_{k \to \infty} x_{k+1}=x_k=t_0=2 
$                             
so we might call $t_0=2$ a "fixpoint" of that iterated operation, and because from all initial values (in a certain interval) it arrives at the same fixpoint, we call it an "attracting" fixpoint.         
For instance            
$ \quad \qquad \displaystyle  x_0=5.3 \\
 \quad \qquad   x_k = b ^{x_{k-1}} \\
 \quad \qquad   \lim_{k \to \infty} x_{k+1}= \infty 
$                      
begins outside that interval and the iteration diverges to $\infty$                     
But if some value is a fixpoint for an operation, then it should also be the fixpoint of the reverse operation, thus we might look at                
$  \quad \qquad \displaystyle x_0=1.1 \\
 \quad  \qquad    x_k = \log_b (x_{k-1}) \qquad \qquad \text{where} \log_b(\cdot) = \log(\cdot)/ \log(b) \text{ and } b=\sqrt2\\
 \quad  \qquad      \lim_{k \to \infty} x_{k+1} =x_k= t_{-1} \qquad \qquad \text{ (=some complex value) }
$                 
But now if you begin at $x_0=5.3$ then we find convergence                 
$ \quad \qquad \displaystyle x_0=5.3 \\
  \quad \qquad    x_k = \log_b (x_{k-1}) \\
 \quad  \qquad      \lim_{k \to \infty} x_{k+1} =x_k= t_1 = 4
$
$\qquad \qquad $ (and the same for any value $2 \lt x_0 \le 4$) 
which is obviously "attracting" for the reverse operation in an interval in which the original operation has a divergent trajectory.          
If in a third view we modify $x_0$ only the slightly least bit around $t_1$ then the original operation runs away from $4$, and so we might introduce the name "$t_1=4$ is a repelling fixpoint".               
The same can now be applied to your other base where you find $b= 3^{1/3}$ . You can insert the exact value $x_0=3$ as initial value and shall find that it is a fixpoint. However it is a repelling one: if you let $x_0 = 3-\epsilon$ then the original operation iterates to the attracting fixpoint which is a bit larger than 2 and if you apply the reverse operation you'll find the repelling fixpoint $t_1 = 3$ which is what you wanted.
