Tetration Cube

Note: $W(\cdot)$ denotes the product log which is the inverse function for $xe^x$ .

$\begin{aligned} ^{3}x &= x^3 \\ \Longrightarrow x^{x^x} &= x^3 \\ \Longrightarrow x^x \ln x &= 3\ln x \\ \Longrightarrow x^x &= 3 \\ \Longrightarrow x &= \sqrt[x]{3} \\ \Longrightarrow x &= e^{\frac{1}{x}\ln 3} \\ \Longrightarrow 1 &= \frac{1}{x}e^{\frac{1}{x}\ln 3} \\ \Longrightarrow \ln 3 &= \frac{1}{x} \ln 3 e^{\frac{1}{x} \ln 3} \end{aligned}$ Here, we can use the product log to solve this:

$\begin{aligned} \ln 3 &= \blue{\ln 3 \frac{1}{x}}e^{\blue{\frac{1}{x}\ln 3}} \\ \Longrightarrow W(\ln 3) &= \blue{\ln 3 \frac{1}{x}} \\ \Longrightarrow \frac{W(\ln 3)}{\ln 3} &= \frac{1}{x} \\ \Longrightarrow x &= \green{\boxed{\frac{\ln 3}{W(\ln 3)}}} = \green{\boxed{e^{W(\ln 3)}}} = \green{\boxed{1.825455022924830040041\dots }} \end{aligned}$

$\begin{aligned} ^3x & = x^3 \\ x^{x^x} & = x^3 \\ \implies x^x & = 3 \end{aligned}$

By numerical method $x \approx \boxed{1.825}$ . (Note that the eventual way of finding the value of $W(\cdot)$ is numerical.)