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 $\log_{10}(x)$?
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 $\log_{10}(x)$?
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 $\ln 10$.