So I continue:

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

So I continue:

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 ? Or are you trying to prove that this statement is true for all ? - 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

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.

Then you end up with

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.