How do I show what values of $N$ that $10N\log N > 2N^2?$

Question

My goal is to find the values of $N$ such that $10N \log N > 2N^2$

I know for a fact this question requires discrete math.

I think the problem revolves around manipulating the logarithm. The thing is, I forgot how to manipulate the logarithm using discrete math.

My question is how do I manipulate this equation in a way such that I can find the values of N such that the equation is true?

Answer 1

First divide by $2N$ on both sides,

$5\log N> N$ (since $N>0$ then the inequality stays the same)

Then by raising to the $e$ power on both sides (the exponential is an increasing function) you'll get

$e^{5\log N}>e^{N}\implies e^{\log N^5}>e^N \implies N^5>e^{N}$

Can you end it from here?

Answer 2

$N \in \mathbb{N}$ ?

$10N\log(N)> 2N^2$

If and only if

$5 \log(N)> N$

If and only if

$e^{5\log(N)} > e^{N}$

If and only if

$N^5 > e^{N}$

Answer 3

In the real domain, consider the function $$f(x)=5\log(x)-x$$ The first derivative cancels at $x=5$ and by the second derivative test, this is a maximum. So, there is a limited range of $x$ where $f(x) >0$.

Sooner or later, you will learn that the zero's of $f(x)$ are given in terms of Lmabert function, that is to say that $f(x) >0$ if $$-5 W\left(-\frac{1}{5}\right) < x < -5 W_{-1}\left(-\frac{1}{5}\right)$$ which, numerically are $1.30$ and $12.71$.

So, for your problem with integer numbers $2 \leq n \leq 12$.