If n = (1999)!,then sumlimits_{x = 1}^{1999} | vedclass

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(A) Given the expression $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ where $n = (1999)!$. Using the property of logarithms $\log_b a + \log_b c = \log_

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$1$

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(A) Given the expression $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ where $n = (1999)!$. Using the property of logarithms $\log_b a + \log_b c = \log_b (ac)$,we can write: $\sum\limits_{x = 1}^{1999} {{\log }_n x} = \log_n 1 + \log_n 2 + \dots + \log_n 1999$ $= \log_n (1 	imes 2 	imes 3 	imes \dots 	imes 1999)$ $= \log_n (1999)!$ Since $n = (1999)!$,the expression becomes $\log_{(1999)!} (1999)!$. Using the property $\log_a a = 1$,we get $1$.

If $n = (1999)!$,then $\sum\limits_{x = 1}^{1999} {{\log }_n x}$ is equal to

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