Hello, I am trying to find out the folowing question:

For what integer values of N is 10N lg N > 2N^2 ?
a) 0 <= N <= 12
b) 1 < N <= 12
c) 0 < N < 12
d) 0 < N < 13

I think the answer is B because you get an error if using a negative number. But, I was hoping that someone might be able to explain how to figure this question out.

One thing I should add is that when I try to plug in a number less than 0, I get an error. When I try a number between 0 and 1 I get this:
10(0.1) (lg 0.1) > 2(0.1)^2
-1 > 0.02
Thus this is why I pick B as my answer.

But how is this related to Java? If you have a Java program which doesn't behave as expected for the above problem, post it, otherwise I would be moving this to the Computer Science sub-forum.

The book I have the question from is "Algorithms in Java". So, I figure that this would be the appropriate place to ask the question. But, we can move this question so that I can get help by others.

You can't take a logarithm of a negative number and it's irrelevant to this problem anyways since N is always positive for all the answers. If N is positive, then so is 2N^2, so the right side of the equation is positive.

For the left side, since N is positive, so is 10N. Logarithms of numbers less than 1 are negative, so logN is negative for N < 1, so when N is less than 1, you are multiplying a positive times a negative and the result is a negative. Again, the right side of the equation is positive. A negative number cannot be greater than a positive number, so you can cross off any answer where N can be less than 1. That leaves only one answer left: B. So you got it right.

Is this some kind of trick question? What is stopping you from trying out the different values of N and seeing the answer?

Hi
I am not that good in Math,,but having a first look at the options i think N>1 it will be the case.. as if N==0 lhs will be 0 and cat be greater than the LHS

correct me if i am wrong