Question

What is
$\frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+\frac{1}{{\log }_{4}N}+......\frac{1}{{\log }_{100}N}$ equal to $(N\ne 1)$?

• A. $\frac{1}{{\log }_{100!}N}$
• B. $\frac{1}{{\log }_{99!}N}$
• C. $\frac{99}{{\log }_{100!}N}$
• D. $\frac{99}{{\log }_{99!}N}$

Answer 1

The correct option is A $\frac{1}{{\log }_{100!}N}$

As ${\log }_{a}x=\frac{1}{{\log }_{x}a}$

$\therefore \frac{1}{{\log }_{2}N}+\frac{1}{{\log }_{3}N}+\frac{1}{{\log }_{4}N}+...+\frac{1}{{\log }_{100}N}={\log }_{N}2+{\log }_{N}3+...+{\log }_{N}100$

$={\log }_N(2.3.\mathrm{4....100})={\log }_N\left(100\right)!=\frac{1}{{\log }_{100!}N}$.