# Show that $k\ln k \in \Theta(n)$ implies $k \in \Theta(n/\ln(n))$?

It is exercise (3.2-8) from Introduction to Algorithms book. I need help to solve it.

I am confused by the fact that there are two parameters. Because usually one parameter is used. There is related exercise

Thanks.

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The question in the title is not related to that exercise: here there is only one parameter $n$, and $k$ is supposed to depend on $n$. – Robert Israel Aug 5 '12 at 17:11
Are you sure? Because if $k$ is function of $n$ then it should be $k(n)$. – Sergey Sokolov Aug 5 '12 at 17:45
If $k$ is not a function of $n$ then $k\ln k$ is a constant with respect to $n$, and thus necessarily in $\Theta(n)$; then it wouldn't make any sense to say something about what this "implies". – joriki Aug 5 '12 at 17:56

If $f \in\Theta(g)$, say $f \sim g$.
So we have $n \sim k\ln k$ and want to show $n/\ln n \sim k$. Well if $n \sim k\ln k$, then $\ln n \sim \ln (k \ln k) \sim \ln k + \ln \ln k \sim \ln k$.
So $n/ \ln n \sim k \ln k / \ln k \sim k$