How log2(n^2) < log3(n^3) when n is large?

#1

I didn’t understand how this comparison is laid in this video, as acc to maths log2(n^2) > log3(n^3) for all n >1. hence binary tree cannot be an efficient way to apply heaps as compared to higher order trees.

#2

@parwandarachit1601
log2(n^2) > log3(n^3) yes you are right,
but the number of comparisons would be more in terms of ternary heaps.
The argument is similar as why binary search is better over ternary search.

#3

but then, it means the argument given in the video is wrong, as the time for binary trees is surely greater than that for ternary trees

#4

yes, i’ll inform the team