Explain $x^{x^{x^{{\cdots}}}} = \,\,3$

Solution 1:

Here's a question: if $x = \sqrt[3]{3}$, then what is the value of $y = x^{x^{x^{\dots}}}$?

As you say, if $y$ is a solution to this, then $y \ln x = \ln y$, so that $$ \frac{\ln y}{y} = \ln(x) = \frac{\ln(3)}{3} $$ Now, how can we "solve this" for $y$? As it ends up, there are two solutions for $y$. The answer you are getting is the second root, $y \approx 2.47805$.

We can, in fact, conclude that the equation $x^{x^{x^{\dots}}} = 3$ has no solution.

So, the salient question is how do we "choose" one value of $y$ over the other?

First of all, we should decide when this sequence converges at all. For a fixed $a$, define the function $f_a(x) = a^x$. Consider the following recursive definition of a sequence:
$$ x_0 = 1\\ x_n = a^{x_{n-1}} $$ If any one value for $y = a^{a^{a^{\dots}}}$ makes sense, it's the limit $\lim_{n \to \infty} x_n$. This has now become an analysis of fixed-point iteration.

It's relatively easy to show that we can guarantee that this sequence converges as long as near $x = 1$ (our starting point), $|f'(x)| < 1$. Because $f'(x) = \ln(a) a^x$, we find that $|f'(1)| < 1$ exactly when $a < e$. So, this sequence will necessarily converge for $a \in (e^{-1},e)$.

As it turns out, however, this is not the only situation in which the sequence converges. In fact, it was shown by Euler (see wiki page and comments below) that the sequence will converge for $e^{-e} < a < e^{1/e}$. Correspondingly, the limit of the sequence will necessarily lie in $(1/e,e)$. Because $3$ lies beyond these bounds, it cannot be the limit of such a sequence.

See also this wikipedia page.

Solution 2:

$$ x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}=y $$ $$ \ln{x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}}=\ln y $$ $$ x^{x^{x^{\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}}}}\ln{x}=\ln y $$ $$ y\ln{x}=\ln y $$ $$ \ln{x}=(\ln{y})/y $$ $$ \ln{x}=\ln({y^{1/y}}) $$ $$ x= y^{1/y} $$

Substitute in $2$ and $3$ to see why these are the solutions of $x$. They will always be the $y^{\text{th}}$ root of $y$.

Solution 3:

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.