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March 31st, 2016, 12:57 AM   #1
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Logarithm feature

Hello,
Is this true? with which formula? Thanks
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March 31st, 2016, 02:48 AM   #2
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$\displaystyle \log{1} + \log{2} + \log{3} + \log{4} + ... + \log{n} = \log{(1 \cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot n)} =\log{(n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot ... \cdot 1)} = \log{n!}$
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March 31st, 2016, 05:21 AM   #3
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The "formula used" is log(a)+ log(b)= log(ab), extended to n terms.
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