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If $n=(1999) !$, then $\sum_{x=1}^{1999} \log _{n} x$ is equal to
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$\sum_{x=1}^{1999} \log _{n} x$
$=\log _{(1999) !} 1+\log _{(1999) !} 2+\ldots+\log _{(1999) !} 1999$
$=\log _{(1999) !}(1.2 .3 \ldots 1999)$
$=\log _{(1999) !}(1999) !=1$
$=\log _{(1999) !} 1+\log _{(1999) !} 2+\ldots+\log _{(1999) !} 1999$
$=\log _{(1999) !}(1.2 .3 \ldots 1999)$
$=\log _{(1999) !}(1999) !=1$
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