Smallest value of n for two algoritms with a certain running time

If one algorithm has a running time of $100n^2$ and another of $2^n$; how can I find the smallest value of $n$ such that the former is faster than the latter?

I could do: $100n^2 < 2^n$ then $\ln(100n^2) < n\ln(2)$ but how do I simplify the left side?

• You try a few values of $n$, get some feeling for what's going on, try a few more values of $n$, and zero in on the answer. Jul 5, 2013 at 12:44
• An explicit form for the intersections of the two graphs is difficult to obtain. An iterative/heuristic approach probably your best shot: W|A link Jul 6, 2013 at 3:23