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

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(C) Using the change of base property of logarithms,$\frac{1}{\log_a b} = \log_b a$.The given expression is $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} +

Vedclass · Accepted Answer

$1$

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(C) Using the change of base property of logarithms,$\frac{1}{\log_a b} = \log_b a$.The given expression is $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dots + \frac{1}{\log_{1983} n}$.Applying the property,this becomes $\log_n 2 + \log_n 3 + \log_n 4 + \dots + \log_n 1983$.Using the product rule for logarithms,$\log_b x + \log_b y = \log_b (xy)$,we get $\log_n (2 	imes 3 	imes 4 	imes \dots 	imes 1983)$.Since $n = 1983!$,the expression simplifies to $\log_n (1983!) = \log_n n$.Since $\log_n n = 1$,the final value is $1$.

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 equal to

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