0 votes 0 votes Evaluate $\frac{1}{n^{\log_{2}n}}$ Quantitative Aptitude logarithms + – rahul sharma 5 asked Sep 12, 2017 rahul sharma 5 445 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes $log Y =log\left ( \frac{1}{n^{log_{2}n}} \right )$ $log Y =log1-log\left ( n^{log_{2}n} \right )$ $log Y =-\left (log_{2} n\right )^{2}$ $Y =2^{-\left (log_{2} n\right )^{2}}$ srestha answered Sep 12, 2017 edited Sep 13, 2017 by srestha srestha comment Share Follow See all 3 Comments See all 3 3 Comments reply A_i_$_h commented Sep 13, 2017 reply Follow Share logY=log(1nlog2n)logY=log(1nlog2n) logY=log1−log(nlog2n)logY=log1−log(nlog2n) logY=−(log2n)2logY=−(log2n)2 Y=e−(log2n)2 // how is this line? 0 votes 0 votes joshi_nitish commented Sep 13, 2017 reply Follow Share @srestha it will be Y = $2^{-(\log_{2}n)^{2}}$ you are using log(base 2) everywhere, then why at last e^ .... 1 votes 1 votes srestha commented Sep 13, 2017 reply Follow Share yah, edited.. 0 votes 0 votes Please log in or register to add a comment.