## Solving n*logn=x and n!=x (with precision) [closed]

Hi, i need help with elementary maths. I don't know, how can i solve equations kind of $n\cdot logn = x$ (for example: $n\cdot logn=10^6$, how can i precisely calculate $n$?). Analogical problem: how yo calculate $n$ in equation $n! = x$? Do I only have Stirling's approximation? Thank's for help.

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How precise do you want to solve this equation? And how advanced tools do you want to use? It's easy to solve $n \cdot \log(n) = 10^6$ using Newton's method with the initial guess $n \approx 10^5$. So, I guess this is not sophisticated enough for you? (Same applies to the second question...) – Helge Jul 22 2010 at 20:35
Of course here solve means "finding a decimal expression for $n$". – Helge Jul 22 2010 at 20:35
For the second question, you want to use the gamma function. Of course these are both examples of solving $f(x)=a$ for a given function $f$ and there are standard numerical methods to do this. – Robin Chapman Jul 22 2010 at 20:40
2d question: If $n!=x$ (you are promised that a preimage of x exists) and you do not care about the running time, the greedy method (divide x by 2, then 3, then 4, etc.) will give you an exact answer. Otherwise your question is about inverting the Gamma function. Or maybe I did not get your question. – Loick Jul 22 2010 at 20:44
I'm afraid this question is not appropriate for MathOverflow. Please see the FAQ for a list of sites where you can get some help. – S. Carnahan Jul 22 2010 at 20:52
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