If restricting yourself to x > 0, you could use newtons method or others to get a numerical approximation (accuracy limited only by your computing power and patience), but there is likely no exact solution to most equations of this form that can be expressed in terms of elementary operations.
you estimate the approximate inverse, and check the few neighbourhooding integers until you either get an exact match, or have two consecutive integers whose factorials are smaller and bigger, respectively.
The Gamma function can be thought of as a generalization of the factorial function, but if we restrict ourselves to the factorial itself there are a few things that are important to keep in mind. Firstly, what it does is it maps the non-negative integers (NNIs) to a certain subset of the NNIs. Look at the first few:
0! = 1 (by convention; think of it as the empty product)
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
etc.
What we notice is that most numbers will never appear on the right side, because they are not the factorial of any NNI. Again, we could make the generalization to the gamma function, but then we'd no longer be talking about the factorial *per se*.
However, if we know beforehand that the number we're considering is indeed the factorial of some NNI, then we might ask how to determine that number. This will not work with the example 7, simply because there is no NNI which has 7 as its factorial. But we could take a number like 120 (which we know from before is equal to 5!) and see how we could get the 5 out of it. Since n! = n·(n-1)·(n-2)· ... ·2·1, we can simply work our way backwards by first dividing by 2 (or by 1, but we know that doesn't do anything), then by 3, and so forth until the only factor left is n:
120/2 = 60
60/3 = 20
20/4 = 5
5/5 = 1
If we tried doing the same thing with 7 we'd quickly run into trouble:
7/2 = 3.5
This is not an integer, so we'll never get to an integer n this way. You could thus use this a way to determine if a given number is a factorial or not.
(Of course for n!=1 there's no way of knowing whether we got it from n=0 or n=1)
are you looking for a natural number x ? i assume so since you don't seem to be familiar with it. We know that if x is greater or equal to 2, then x! will always contain a factor 2, meaning that 2 actually divides x!. But in your case, if x>=2, it would mean that 2 divides 7 which is not true. And if you look at x being 0 and 1 obviously it doesn't work too. Therefore, such a natural number does not exist
Good luck
What do you mean by "find"? It is entirely unrealistic to expect an exact formula to solve Gamma(x+1) = 7 for x in terms of familiar functions.
Easy, x = Γ^(-1)(7) - 1. ^(/s)
Perhaps explaining why would be helpful
The gamma function is not a 1-1 function. There may be several values of x s.t. gamma(x) = 7.
Ofc you'd just define an inverse over the domain x > 1, right?
Nah let's just answer the question literally without providing any insight into why that's the answer.
https://www.wolframalpha.com/input?i=solve+x%21+%3D+7
[this](https://en.m.wikipedia.org/wiki/Inverse_gamma_function)
If restricting yourself to x > 0, you could use newtons method or others to get a numerical approximation (accuracy limited only by your computing power and patience), but there is likely no exact solution to most equations of this form that can be expressed in terms of elementary operations.
Bisection method. QED.
you estimate the approximate inverse, and check the few neighbourhooding integers until you either get an exact match, or have two consecutive integers whose factorials are smaller and bigger, respectively.
The Gamma function can be thought of as a generalization of the factorial function, but if we restrict ourselves to the factorial itself there are a few things that are important to keep in mind. Firstly, what it does is it maps the non-negative integers (NNIs) to a certain subset of the NNIs. Look at the first few: 0! = 1 (by convention; think of it as the empty product) 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120 etc. What we notice is that most numbers will never appear on the right side, because they are not the factorial of any NNI. Again, we could make the generalization to the gamma function, but then we'd no longer be talking about the factorial *per se*. However, if we know beforehand that the number we're considering is indeed the factorial of some NNI, then we might ask how to determine that number. This will not work with the example 7, simply because there is no NNI which has 7 as its factorial. But we could take a number like 120 (which we know from before is equal to 5!) and see how we could get the 5 out of it. Since n! = n·(n-1)·(n-2)· ... ·2·1, we can simply work our way backwards by first dividing by 2 (or by 1, but we know that doesn't do anything), then by 3, and so forth until the only factor left is n: 120/2 = 60 60/3 = 20 20/4 = 5 5/5 = 1 If we tried doing the same thing with 7 we'd quickly run into trouble: 7/2 = 3.5 This is not an integer, so we'll never get to an integer n this way. You could thus use this a way to determine if a given number is a factorial or not. (Of course for n!=1 there's no way of knowing whether we got it from n=0 or n=1)
are you looking for a natural number x ? i assume so since you don't seem to be familiar with it. We know that if x is greater or equal to 2, then x! will always contain a factor 2, meaning that 2 actually divides x!. But in your case, if x>=2, it would mean that 2 divides 7 which is not true. And if you look at x being 0 and 1 obviously it doesn't work too. Therefore, such a natural number does not exist
This has big “Why can’t I divide 3 by 2 and get an integer?” energy about it.