How to solve $x^{x^{x^x}} = 1/3^{\sqrt{48}}$ How to solve
$$x^{x^{x^x}} = \frac{1}{3^{\sqrt{48}}}$$
Attempt :
Let $x^{x^{\cdots}} = y$
$$\begin{align}
x^y &=y\\
y\ln(x) &= \ln(y)\\
-\ln(x) &= -\ln(y)e^{-\ln(y)}\\
-\ln(y) &= W(-\ln(x))\\
y &= e^{-W(-\ln(x))}
\end{align}$$
I'll stop this. Am i doing this right? I know the original question is not continuous power. By the way i'm assuming $x^{x^{x^x}} = x^{x^{\cdots}}$ But is this allowed?
Some advice and help are needed.
Thanks
 A: There is no real-valued solution.  We first show that for $x > 0$, $x^x \ge x$.  If $x \ge 1$, then $x^x \ge x^1 = x$.  If $0 < x < 1$, then $$\log (x^x) = x \log x > 1 \log x = \log x,$$ since $\log x < 0$.  Therefore, $$x^{x^{x^x}} \ge x^{x^x} \ge x^x \ge x.$$  Next, we show $x^x \ge 1/2$.  For the reason above, we need only consider $0 < x < 1$.  Then by differentiation, if $f(x) = x^x$, then $$f'(x) = f(x) (\log f(x))' = x^x (1 + \log x),$$ and since we established that $f(x) \ge x$ for $0 < x < 1$ the only critical point is when $x = e^{-1}$.  The second derivative $$f''(x) = x^x \left(x^{-1} + (1 + \log x)^2\right)$$ shows that $f$ is a convex function on the same interval.  Thus $x = e^{-1}$ is the absolute minimum and the minimum value attained is $f(e^{-1}) = e^{-1/e} > 1/2$.  Consequently $$x^{x^{x^x}} > x^{x^{1/2}} = \left((x^{1/2})^{x^{1/2}}\right)^2 > (1/2)^2 > 1/4 > 3^{-4 \sqrt{3}}.$$
A: As said in comment, in the real domain, there is no zero for the function
$$f(x)=x^{x^{x^x}}-3^{-\sqrt{48}}$$ the first derivative
$$f'(x)=x^{x^x+x^{x^x}-1} \left(x^x \log (x) (x \log (x) (\log (x)+1)+1)+1\right)$$ cancels close to $x_*\sim 0.275$ (this is a minimum by the second erivative test) and $f(x_*) \sim 0.593$.
If the problem was
$$g(x)=x^{x^{x^x}}-3^{\sqrt{48}}$$ it would be a very different story. Plotting
$$h(x)=\log \left(\log \left(\log \left(x^{x^{x^x}}\right)\right)\right)-\log \left(\log
   \left(4 \sqrt{3} \log (3)\right)\right)$$ shows almost a straight line around $x=2$.
Newton method will work like a charm
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 2.0000000 \\
 1 & 1.9447990 \\
 2 & 1.9466308 \\
 3 & 1.9466333
\end{array}
\right)$$
Edit
Back to the original equation, there are at least two complex roots which are
$$x_\pm=-0.332844\pm 0.291254\, i$$
In comments, I have asked how I found these roots. In a preliminary step, I looked at function
$$F(a)=\Im(f(-a(1+i)))^2+\Re(f(-a(1+i)))^2$$ and noticed that for $a \sim \frac \pi {10}$ the result was very small $(F\left(\frac{\pi }{10}\right)\sim  1.37 \times 10^{-6}$).
Now, Newton iterations
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & -0.31415927-0.31415927\, i \\
 1 & -0.32787054-0.30029210\, i \\
 2 & -0.33258317-0.29293379\, i \\
 3 & -0.33287291-0.29129953\, i \\
 4 & -0.33284380-0.29125412\, i \\
 5 & -0.33284375-0.29125414\, i
\end{array}
\right)$$
