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Suppose that I had to find $\log_{10}(8952!)$. Now, since $\log(a) + \log(b) = \log(ab)$, this can be rewritten to the following summation:

$$\sum_{x=1}^{8952}{\log_{10}(x)}$$

Would there be a general way to compute this in terms of $\log_{10}(x)$?

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Did you try wolfram Alpha's computational engine? –  Mohamed Ennahdi El Idrissi May 25 at 22:05
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1 Answer 1

Stirling's approximation gives a pretty tight bound for the natural logarithm of $n!$ (but not an exact value), and then the base-10 logarithm is only a matter of dividing by $\ln 10$.

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Nice answer. It is also possible to get even better approximations using Euler-Maclaurin formula i think. Not ? –  mick Sep 9 '12 at 20:43
    
There are lots of versions of Stirling's formula. Euler-Maclaurin for $\ln(n!)$ corresponds to one of them. –  Robert Israel Sep 9 '12 at 21:12
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