If $n = (2020)!$,then the value of $\frac{1}{\log _{2} n} + \frac{1}{\log _{3} n} + \frac{1}{\log _{4} n} + \ldots + \frac{1}{\log _{2020} n}$ is equal to:

  • A
    $2020$
  • B
    $1$
  • C
    $(2020)!$
  • D
    $0$

Explore More

Similar Questions

The number of solutions of the equation $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$ is

If $A = \log_2 \log_2 \log_4 256 + 2 \log_{\sqrt{2}} 2$,then the value of $A$ is:

What is the number of solutions for the equation $\log_4(x - 1) = \log_2(x - 3)$?

Difficult
View Solution

If $\cosh ^{-1}\left(\frac{5}{3}\right)+\sinh ^{-1}\left(\frac{3}{4}\right)=k$, then $e^k=$

Which one of the following observations is correct for the features of the logarithm function to any base $b > 1$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo