Question

If $\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+...+\frac{1}{{\log }_{53}n}=\frac{1}{{\log }_{k!}n}$, then find the value of $k$, $(n>1)$

• A. $52$
• B. $53$
• C. $54$
• D. $n+53$

Answer 1

The correct option is B $53$

$\frac{1}{{\log }_{2}n}+\frac{1}{{\log }_{3}n}+...+\frac{1}{{\log }_{53}n}$
$={\log }_{n}2+{\log }_{n}3+...+{\log }_{n}53,[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$={\log }_n(2\times 3\times ...\times 53)={\log }_{n}53!=\frac{1}{{\log }_{53!}n}$
$\therefore k=53$