\left. \begin{array} { l } { \text { If } n = ( 2017 ) ! , \text { the

1.	\left. \begin{array} { l } { \text { If } n = ( 2017 ) ! , \text { then what is } } \\ { \frac { 1 } { \operatorname { log } _ { 2 } n } + \frac { 1 } { \operatorname { log } _ { 3 } n } + \frac { 1 } { \operatorname { log } _ { 4 } n } + \ldots + \frac { 1 } { \operatorname { log } _ { 2017 } n } } \end{array} \right.
Answer» n = (2017)! use the identity loga[b] = logb/loga so, log2[n] = logn/log2 and 1/log2[n] = 1/(logn/log2)=> 1/log2[n] = log2/logn similarly for all other terms 1/log3[n] = log3/logn1/log4[n] = log4/logn now adding all we will get , denominator as same in all as logn on adding= (log2+log3+log4 +......+log2017)/(logn)= log(2×3×4×....2017)/log(n)= log(2017!)/logn= logn/logn = 1

\left. \begin{array} { l } { \text { If } n = ( 2017 ) ! , \text { then what is } } \\ { \frac { 1 } { \operatorname { log } _ { 2 } n } + \frac { 1 } { \operatorname { log } _ { 3 } n } + \frac { 1 } { \operatorname { log } _ { 4 } n } + \ldots + \frac { 1 } { \operatorname { log } _ { 2017 } n } } \end{array} \right.

Answer»

n = (2017)!

use the identity loga[b] = logb/loga

so, log2[n] = logn/log2 and 1/log2[n] = 1/(logn/log2)=> 1/log2[n] = log2/logn

similarly for all other terms 1/log3[n] = log3/logn1/log4[n] = log4/logn

now adding all we will get , denominator as same in all as logn

on adding= (log2+log3+log4 +......+log2017)/(logn)= log(2×3×4×....2017)/log(n)= log(2017!)/logn= logn/logn = 1