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I've got $8n^2 \lt 64n\log(n)$ and I need to find the $n$ range if $n\gt 0$ to satisfy the inequality.

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2 Answers 2

up vote 4 down vote accepted

So this is n<8*ln(n). If you plot it you can find a range of n that works quite easily. You probably won't be able to solve it exactly, but approximate numerical solutions are available. I find a range of about 1.2 to 26.

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Wolfram Alpha gives a range of approximately 1.15537 to 26.0935. –  Joshua Shane Liberman Jan 26 '11 at 17:42
There's no other way to solve an inequality but to plot the graph? –  Denys S. Jan 26 '11 at 17:44
Oops, I got a base 10 log, not natural log. I'll edit to correct –  Ross Millikan Jan 26 '11 at 17:45
@den-javamaniac: Without the Lambert W function, there is no algebraic answer. So all you can do is approximate it numerically. You can do that with a graph or with root finding. –  Ross Millikan Jan 26 '11 at 17:47

$n$ suggests that it might be a natural number.

Therefore you could use

$8 n^2 < 64 n log(n)$

$\Longleftrightarrow 0 < 8 log(n) - n$

You can easily see that $n$ grows much faster than $log(n)$. Therefore you can see that for $n > 26$ there are no integer solutions. Also the case $n=1$ is problematic as $log(1)=0$. So all in all the numbers $n = \{2,3,\ldots,26\}$ are a solution for your problem. If you need real numbers you will have to use, like already mentioned the Lambert W function ( http://en.wikipedia.org/wiki/Lambert_W_function )

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+1 for Lambert W function) –  Denys S. Jan 27 '11 at 14:21

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