1 votes 1 votes Order the following: $\frac{e^{n \log n}}{n}, 2^{n \log n}, n^{\sqrt n}$ Algorithms made-easy-test-series asymptotic-notation + – shweta sah asked Dec 2, 2018 • retagged Jun 10, 2022 by makhdoom ghaya shweta sah 399 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Magma commented Dec 2, 2018 reply Follow Share $\frac{e^{n logn}}{n} = \frac{n^{n}}{n} = n^{n-1}$ $n^{n^{1/2}}$ $2^{nlogn} = n^{nlog 2}$ Option C is correct 2 votes 2 votes chauhansunil20th commented Dec 2, 2018 reply Follow Share how did you compare n^(n-1) and n^n*log2? log2 will be less than 1 if we assume log base is e as you took for first complexity. 0 votes 0 votes himgta commented Dec 2, 2018 reply Follow Share @Magma What should be the approach to solve such type of questions...I got stuck sometimes! 0 votes 0 votes sudoankit commented Dec 2, 2018 reply Follow Share @himgta check this out, http://bigocheatsheet.com 0 votes 0 votes Please log in or register to add a comment.