Variations on $x^x=y$

Question

Pretty sure at least a few people know how to solve for x in an equation of the form $x^x=y$.

One way (with 2 as an example): $x^x=2$

$x=2^{1/x}$

$x=({e^{ln(2)}})^{1/x}$

Multiplying both sides by $ln(2)/x$:

$ln(2)=ln(2)/x*{e^{{ln(2)}/x}}$

Which can then be solved using the Lambert W function on both sides (I don't understand it completely but I believe it to be something like the inverse of $f(z)=z*e^z$) to yield:

$W(ln(2))=ln(2)/x$

And so $x=ln(2)/W(ln(2))$.

That makes sense, but it relies on the base and exponent being equal to utilise the Lambert W function. Say I were to attempt something similar using $x^{6*x^{12}}=2$, the same process doesn't work exactly as it doesn't fit the form required (of $x*e^x$), even though there's still only one variable. Any ideas on how to get it into this form? Or perhaps use another method entirely? I would prefer if the answer is in exact form, so any iterating equations or graphical methods wouldn't help much.

Thanks and much appreciated :)

Answer 1

The general solution of $x^{ax^b}= c$ is $$x = \left(\dfrac{b}{a} \log(c)\cdot \dfrac1{ W\left(\frac{b}{a} \log(c)\right)}\right)^{1/b}$$

Answer 2

Square $x^{6 x^{12}} = 2$ to get $x^{12 x^{12}} = 4$. Can you take it from here?