I have a function that we use for some game calculations, but now I need to reverse it and solve for x. I have no idea how to do that. The function looks like this:
$$y = 10 + \frac{ ( {ln(x)} ^ {ln(x)} ) ^ {1.2} }{3}$$
I get about as far as this:
$$(3y - 30)^\frac{1}{1.2} = {ln(x)} ^ {ln(x)}$$
And then I'm stumped. How can I solve for x in this scenario?
The inverse function of $f(x) = x^x$ is $$f^{-1}(x) = \frac{\ln x}{W (\ln x)},$$ where $W$ is the Lambert omega function. This inverse is good only for $x>1/e.$
(See https://en.wikipedia.org/wiki/Lambert_W_function.)
So in your equation, you can solve for $\ln x$:
$$\ln x = \frac{\ln( (3y-30)^{1/1.2} )}{W (\ln( (3y-30)^{1/1.2} ))} = \frac{(5/6)\ln( (3y-30) )}{W ((5/6)\ln( (3y-30))}.$$
So $x = e$ raised to that mess.
If this is for some computational purpose, there are series approximations for $W$.