# Which grows asymptotically faster, $\log \sqrt{n}$ or $4 \log n$?

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.

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

## 1 Answer

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 $$\Theta\left(\log ( n^{1/2} ) \right) = \Theta\left( \frac{1}{2}\log n \right) = \Theta\left(\log n \right).$$

• 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 $f(n)=O(g(n))$. Commented Mar 1, 2021 at 10:39