0 votes 0 votes How to find log of the first n natural numbers? $T(n)= \log 1+ \log 2+ \log 3+\ldots + \log n$ // How to proceed from here? Quantitative Aptitude logarithms simplification + – iarnav asked Jan 11, 2018 • recategorized Jun 17, 2022 by Arjun iarnav 622 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Anu007 commented Jan 11, 2018 reply Follow Share log ($ 1\times2 \times3\times4\times......\times n$) = log(n!) = log(nn) = O(nlogn) 3 votes 3 votes hs_yadav commented Jan 11, 2018 reply Follow Share i think n! can be maped to exponential ....nn ... so O(nlogn) 1 votes 1 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes you can write log( 1 *2 *3 * ------------------n) log( n ! ) answer product of 1st n natural number = n ! other one : 1 + 2+ 3+ -------------+ n = n(n+1) /2 same way log 1 + log 2+ -------- log n = $\frac{logn(logn +1 )}{2}$ { but it is not true } sumit goyal 1 answered Jan 11, 2018 • selected Jan 11, 2018 by iarnav sumit goyal 1 comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments iarnav commented Jan 11, 2018 reply Follow Share @sumit goyal 1 Anu007 how you you write log1+ log2+ log3+⋯+ logn as log( 1 *2 *3 * ------------------n) as log( n ! ) 1 votes 1 votes sumit goyal 1 commented Jan 11, 2018 reply Follow Share hi bhai by log property log a + log b = log ( a.b) so if you have , log a + log b + log c +------- + log n then it is equal to log(a . b . c ------- . n ) and n! = 1.2.3.4............ n @iarnav 1 votes 1 votes iarnav commented Jan 11, 2018 reply Follow Share @sumit goyal 1 Thanks Bhaijaan :) 1 votes 1 votes Please log in or register to add a comment.