I have been looking at the question as to which grows faster asymptotically; $\log \sqrt n$ or $4 \log n$. I have applied L'Hopitals rule and ended up with 1/8. This would imply that they grow at same rate.

Graphically $4 \log n$ is always above $\log \sqrt n$. Also from a perspective that $4 \log n$ is $\log n^4$ would imply intuitively that growth of $4 \log n$ is faster asymptotically

There appear to be contradictions.

  • $\begingroup$ Also, 1/2 logn vs 4 logn they would appear to be same growth rate $\endgroup$
    – david
    Commented Mar 1, 2021 at 10:21

1 Answer 1


The asymptotic growth of $4 \log n$ is referred to as $\Theta(\log n)$. You will have to look at the definition of asymptotic growth to see why that is the case, but intuitively, it is the growth of a function when we discard constant factors and only look at the function "in the limit".

You have found out that the difference between the two functions "is 1/8", which makes sense, and which would put these two functions in the same "growth class".

When it comes to $\Theta\left(\log ( \sqrt n ) \right)$, it is, as you likely have found out $$\Theta\left(\log ( n^{1/2} ) \right) = \Theta\left( \frac{1}{2}\log n \right) = \Theta\left(\log n \right).$$

  • 1
    $\begingroup$ I replaced $O(\cdot)$ with $\Theta(\cdot)$ since $f(n),g(n)=O(h(n))$ does not tell us anything about whether $f(n) = \Omega(g(n))$ and/or $f(n)=O(g(n))$. $\endgroup$
    – Steven
    Commented Mar 1, 2021 at 10:39

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