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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 wol$f$ram Alpha's computational engine?2014-05-25

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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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    There are lots of versions of Stirling's formula. Euler-Maclaurin for $\ln(n!)$ corresponds to one of them.2012-09-09