0 votes 0 votes (logn)^k<=cn=logn<=c(n)^ 1/k how?? Algorithms asymptotic-notation time-complexity descriptive + – Raghav Khajuria asked Sep 2, 2018 retagged Jun 23, 2022 by Lakshman Bhaiya Raghav Khajuria 375 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Lets say n=2m Also assume log as log2 A=(lgn)k=(lg(2m))k=mk B=cn=c*2m C=lgn=lg(2m)=m D= c(n) 1/k=2m/k C<=A<=D<=B sakharam answered Sep 2, 2018 sakharam comment Share Follow See all 5 Comments See all 5 5 Comments reply Raghav Khajuria commented Sep 3, 2018 i edited by Raghav Khajuria Sep 3, 2018 reply Follow Share Actually i want to know from (logn) ^k<=cn How can we write this As logn<=c(n) ^1/k K is a constant 1 votes 1 votes sakharam commented Sep 3, 2018 i edited by sakharam Sep 3, 2018 reply Follow Share I am assuming k is a constant with high value. If k=1 or 2 then what I wrote wont be true. Your question does not specify what k is, constant or a variable. So u may consider as a bad question. 0 votes 0 votes Raghav Khajuria commented Sep 4, 2018 reply Follow Share Thanks @saksham for giving explanation ,I got it.. 0 votes 0 votes Raghav Khajuria commented Sep 4, 2018 reply Follow Share I think if we take log both sides then it would be clear which one is greater .. 0 votes 0 votes sakharam commented Sep 4, 2018 reply Follow Share Yes, actually the questions should specify what k is Because for smaller value of k the result I showed need not be true Yes Log works an done more thing u can try is google any online function plotting website and plot functions. 0 votes 0 votes Please log in or register to add a comment.