If n = 1983!,then the value of the expression | vedclass

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(A) Using the change of base formula,$\frac{1}{\log_a n} = \log_n a$. Substituting this into the expression,we get: $\log_n 2 + \log_n 3 + \log_n 4 +

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$1$

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(A) Using the change of base formula,$\frac{1}{\log_a n} = \log_n a$. Substituting this into the expression,we get: $\log_n 2 + \log_n 3 + \log_n 4 + \dots + \log_n 1983$. Using the property $\log_b x + \log_b y = \log_b (xy)$,the expression becomes: $\log_n (2 	imes 3 	imes 4 	imes \dots 	imes 1983)$. Since $n = 1983! = 1 	imes 2 	imes 3 	imes \dots 	imes 1983$,the expression is: $\log_n (1983!) = \log_n n = 1$.

If $n = 1983!$,then the value of the expression $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dots + \frac{1}{\log_{1983} n}$ is:

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