I was reading a book which gave the running time of an algorithm $A$ to be $8n^2$ and another algorithm $B$ to be $64n\log_2{n}$ and asked what is the greatest input size $n$ where $n \in \mathbb{N}$ such that $A$ is faster than $B$ (It is earlier established that $B$ is faster for all values past a certain point $c$, finding that $c$ is the question.)
The problem can be mathematically formulated as solving the following inequality:
$$\tag{1}\quad 8n^2 < 64n\log_2{n}.$$
I did this easily enough by writing a quick program in C (The answer is for all input sizes less than 44), but I was displeased with that method; it just felt dirty, ya know? I'm looking for an analytic solution, should one exist. I've done the following work.
$$8n^2 < 64n\log_2{n}\implies n < 8\log_2{n}\implies n < \log_2{n^8}\implies 2^{n} < n^8.$$
Anyone know how to analytically derive the result of $n<44?$