The Easy Factlog

$log_{a}b =1/ log_{b}a$

Hence the given expression becomes:

$log_{n}2+log_{n}3+......log_{n}2006$

$= log_{n}(2.3.4....2006)$

$=log_{n}n = 1$

1/log(n,2)+1/log(n.3)+..........+1/log(n,2006) =>log(2,n)+log(3,n)+......log(2006,n)

=>log((2 x 3 x 4 x.....x 2006),n) =log(2006!,n)

Given that n=2006!

by substituting n value as 2006! in log(2006!,n)

We get log(2006!,1)=log(2006!,2006!)=1

therefore 1/log(n,2)+1/log(n.3)+..........+1/log(n,2006) when n=2006! is 1

its a property that 1/log to the base a of b is equal to log to the base b of a thus solving we get log to the base n of n that is equal to 1