How to determine $10^{\log n}$ and $3n^2$ which grows faster asymptotically?

Question

My think is pretty easy that $10^{\log n} = n$, which is growing slower than $3n^2$.

However, many tutorial shows that $3n^2$ ranks before $10^{\log n}$.

I'm really confused.

Answer 1

You have to be careful here, since the answer depends on the particular log function you use. As Lieuwe noted if $\log$ in this context means $\log_{10}$ then $10^{\log n}=n$ and certainly $n$ "ranks before" $3n^2$, under any reasonable interpretation of "ranks before". However, if we have a different base for the logarithm, that might not be the case.

It's not hard to show that $10^{\log_b n}=n^{log_b{10}}$ (take the log of both sides) and so $10^{\log_b{n}}$ will be asymptotically larger than $n^2$ as long as $\log_b10>2$, i.e., when $b^2<10$, so when you use logs to base $b$ with, say, $b=3$ you'll have $10^{\log_b n}$ "ranks after" $3n^2$.

Answer 2

If you are using base 10 log, then yes, $10^{\log n}=n$. It has lower asymptotic growth than $3n^2$, as you note, because for every $c$, there is an $n_0$ such that $3n^2>cn$ for all $n$ which are greater than $n_0$. Proving this is an exercise for the reader.

If you read that $3n^2$ ranks before $10^{\log n}$, then those sources are wrong. There are other functions that look superficially similar that do grow faster, such as $n^{\log n} = 10^{(\log n)^2}$ and $\log n^{\log n}=10^{\log n \cdot \log\log n}$.

Answer 3

As others have noted, it depends on the base of the $log$. When a base is not given, in computer science it is generally assumed to be 2. In mathematics, it is generally assumed to be $e$. In engineering, it is generally assumed to be 10.