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?
On
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 $$h(x)=\log(\log(x^{x^x}))-\log(\log(3))$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & 2.0000000 \\ 1 & 1.6165932 \\ 2 & 1.6349681 \\ 3 & 1.6350785 \end{array} \right)$$
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}