# How to solve equations in the form $f(x) = x!$?

Preliminary Google's grant nothing, so I'll pose it here:

How may I go about solving equations for the independent when in the form $f(x) = x!$ ?

$x!$ being the factorial of $x$, i.e.: $4! = 4 \times 3 \times 2 \times 1$

I'm aware of factorial approximation methods such as Stirling's approximation; $x! \approx \sqrt{2 \pi x} \cdot \left(\dfrac{x}{e}\right)^x$; but am aware of no such method to solve for $x$ itself in the natural form.

Thoughts?

• see here – user66081 Aug 21 '16 at 8:05

The factorial extends continuously onto the real numbers via a function called the gamma function, which remarkably has the same values at the integers as the factorial does with a small shift : $\Gamma(n) = (n-1)!$. This permits us to not only look for integer, but also real number solutions for this problem. I suggest you look up this remarkable function on Wikipedia.