Solution of $x^{x^x}=3$ Is it possible to analytically solve the expression $x^{x^x}=3$?
And how to solve this problem numerically?
Simply testing some results, I found that $\sqrt{2.6}<x<\sqrt{2.7}$. What matches the result provided by Elliot G in the comments.
But in addition to a numerical solution, like the technique posted by glowstonetrees, is it possible to find a "closed" formula for the solution of this problem?
 A: Quite sure you can't solve this analytically.
On the other hand, there are many numerical methods for solving $f(x)=0$. For example, Newton's method gives the sequence of iterates
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
for any starting guess $x_0$ that is close enough to the desired solution.
In your case, you would have
$$f(x) = x^{x^x}-3 \qquad \qquad f'(x) = x^{x^x}x^x\bigg( \frac 1x + \ln (x) \big(1+\ln(x)\big)\bigg)$$
So for example you could run a for-loop
\begin{align}
& x_0 = 1 \\
& \text{for } n = 0,1,2,\dots \\
& \; \; \; \; \; x_{n+1} = x_n - \frac{x_n^{x_n^{x_n}}-3}{x_n^{x_n^{x_n}}x_n^{x_n}\Big( \frac {1}{x_n} + \ln (x_n) \big(1+\ln(x_n)\big)\Big)} \\
& \text{end}
\end{align}
A: For sure, a numerical solution could always be obtained using, as already suggested in comments, Newton method which will be the simplest.
You just need to reformulate the problem as : find the zero of function
$$f(x)=x^{x^x}-3$$ The problem is that the function is so stiff that, if you do not have a good estimate, many iterations could be required. For example, let us start with $x_0=2$ which looks to be very close to the solution. The iterates will be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 2.0000000 \\
 1 & 1.8786299 \\
 2 & 1.7574536 \\
 3 & 1.6696436 \\
 4 & 1.6380522 \\
 5 & 1.6351011 \\
 6 & 1.6350785
\end{array}
\right)$$
Trying to make the problem more linear, trying with
$$g(x)=\log(x^{x^x})-\log(3)$$
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 2.0000000 \\
 1 & 1.7499438 \\
 2 & 1.6481903 \\
 3 & 1.6352591 \\
 4 & 1.6350785
\end{array}
\right)$$ One more step in the same direction with
$$h(x)=\log(\log(x^{x^x}))-\log(\log(3))$$
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 2.0000000 \\
 1 & 1.6165932 \\
 2 & 1.6349681 \\
 3 & 1.6350785
\end{array}
\right)$$
