The equation for Stirling's Approximation is the following:
$$x! = \sqrt{2\pi x} * (\frac{x}{e})^x$$
Writing as a function for y gives us the following:
$$y = \sqrt{2\pi x} * (\frac{x}{e})^x$$
Is there a way to solve this equation for x, effectively finding an inverse to this function?
Instead of the inverse to Stirling's approximation, you can (and should) consicder the inverse to the Gamma function itself. See this MathOverflow discussion.
For the inverse of the factorial function, @Gary proposed in year $2013$ a superb approximation (have a look here)
$$x!= y \quad \implies \quad x \sim \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} - \frac{1}{2}\tag 1$$ where $W(.)$ is Lambert function.
If you want to make a less good approximation by yourself, start taking the logarithms of both sides and use Stirling approximation. This gives $$\log(y)=x (\log (x)-1)+\frac{1}{2} \log (2 \pi )+\frac{1}{2} \log ( x)+O\left(\frac{1}{x}\right)$$ Ignoring the last term $$\log(y)=x (\log (x)-1)+\frac{1}{2} \log (2 \pi )\implies x= \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}}\tag 2$$
Using for example $y=120$, $(1)$ would give $x=4.99556$ while $(2)$ would give $x=5.49556$, while, as you know, the exact solution is $5$.
