$\displaystyle 8n^2 < 64n* log_2 n$
So I continue:
$\displaystyle n< 8* log_2 n$
$\displaystyle n< log_2 n^8$
$\displaystyle 2^n < n^8$
Not sure if this is correct procedure at all. Anyway, I'm stuck here...
I'm trying to find the values of n, where the statement is true. So the answer would be in the form: n < some number
I did set up a table in excel. I know the answer is between 43 and 44. That is for n= 43, 2^n is smaller. But for n=44, n^8 is smallest.
But would like to know how the procedure is to get to that result.
There are two solutions for $\displaystyle 2^n<n^8$
One solution involves negative n.
If the solution is for positive n, since your original equation is a logarithm,
you could use a change of base to the natural logarithm.
This makes a graphical solution much easier.
$\displaystyle \displaystyle\huge\log_2n=\frac{log_en}{log_e2}$
$\displaystyle log_e2=ln2=constant$
Then you end up with
$\displaystyle n<\frac{8}{log_e2}log_en$
$\displaystyle \frac{8}{log_e2}log_en-n>0$
You can use the Newton-Raphson method to find the two "n" for which this is zero.
Your inequality holds for "n" between those two values.
Thank you! I think it will suffice to just do the graph on my calculator when reaching an answer. It is computer science
and Newton-Raphson method is outside scope of the course I am taking. The question was for what input size n one algorithm
ran faster the the other one.