I found a small exercise which I couldn't figure what to do, so I found a solution. Then I tried to understand it and everything went well until I got to this part:
$$\frac{1}{8} = \frac{\log(n)}{n}$$
Then it just skipped and say that the answer was $n = 43$. I was wondering if there is some kind of property for $\log (n) / n$ I don't know about. Or otherwise, how was this solved?
EDIT: This is the exercise
Suppose we are comparing implementations of insertion sort and merge sort on the
same machine. For inputs of size n, insertion sort runs in 8n2 steps, while merge
sort runs in 64n lg n steps. For which values of n does insertion sort beat merge
sort?
And this was the solution given:
$$8n^2 = 64n \log(n)$$
$$n^2 = 8n \log(n)$$
$$n = 8 \log(n)$$
$$\frac{1}{8} = \frac{log(n)}{n}$$
Best Answer
If we take your amended question, which asks us to solve the inequality:
$$8n^{2}<64 n \log_{2}(n)$$
We can therefore divide both sides by $64n$ to get:
$$\frac{n}{8}<\log_{2}(n)$$
This has to be solved by numerical methods and we see that we have (by Mathematica):
$$1.1<n\lesssim43.5593$$
And we note that $n$ must be an integer so we have:
$$\lceil 1.1 \rceil=2 <n < 43 = \lfloor 43.5593\rfloor$$