$\displaystyle 8n^2 < 64n* log_2 n$

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...

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- Aug 14th 2010, 07:18 AMLars91Inequality with logarithm, help needed
$\displaystyle 8n^2 < 64n* log_2 n$

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... - Aug 14th 2010, 07:23 AMProve It
What are you trying to do here?

Are you trying to solve this inequality for $\displaystyle n$? Or are you trying to prove that this statement is true for all $\displaystyle n$? - Aug 14th 2010, 07:39 AMLars91
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. - Aug 14th 2010, 11:17 AMArchie Meade
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. - Aug 15th 2010, 05:15 AMLars91
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.