I was recently looking at the article on the P complexity class on Wikipedia. In the section on relationships to other classes, it mentions that P is known to be at least as large as L, giving this explanation:
A decider using $O(\log n)$ space cannot use more than $2^{O(\log n)} = n^{O(1)}$ time.
The article sadly fails to explain how that last equality holds. How exactly does one go about showing that $2^{O(\log n)} = n^{O(1)}$?