proving " if $k\log(k) = \theta (n)$ then $k = \theta (\frac{n}{\log(n)})$"
First, let me explain why your claim holds. Then we'll see how to prove it.

If $k \log k = \Theta(n)$ then $k$ has the same order of magnitude as $n$, and so we expect $\log k \approx \log n$ (or, more formally, $\log k = \Theta(\log n)$). Diving by $\log n$, we deduce $k = \Theta(n/\log n)$.

This was just intuition, so now let us see a rigorous proof. A rigorous proof rests on a rigorous definition of big $\Theta$, and this is a surprisingly delicate matter. In this case it suffices to assume that $n \geq 1$.

Upper bound

Suppose that $k \log k \leq Cn$ for $C&gt;0$.

If $k \leq \sqrt{n}$ then
$$
\frac{k}{\frac{n}{\log n}} \leq \frac{\log n}{\sqrt{n}}.
$$
Now $\frac{\log n}{\sqrt{n}}$ is continuous and tends to zero at infinity, so it is upper-bounded by some constant $M$. Therefore in this case
$$
k \leq M \frac{n}{\log n}.
$$

If $k &gt; \sqrt{n}$ then
$$
Cn \geq k \log k &gt; k \log \sqrt{n} = \frac{1}{2} k \log n,
$$
and so
$$
k \leq 2C \frac{n}{\log n}.
$$

We conclude that for all $k$,
$$
k \leq \max(2C,M) \frac{n}{\log n}.
$$

Lower bound

Left to you as an exercise.