Theta Manipulation to show $N = \Theta(n/\log N)=\Theta(n/\log n)$ I am studying different models of computation and how algorithms can be interpreted under different models.

Here is a math(?) question that has been bugging me.

Suppose we have $n = \Theta(N\log N)$ bits as the size of input and there are N elements in the input. (The implied assumption is each element has $\Theta(\log N)$ bits).

Now we know Insertion Sort has running time of $\Theta(N^2)$ in the worst case with respect to N, the number of elements in the input.

Under the RAM model, the input is $n = \Theta(N\log N)$ bits. So we can say $N = \Theta\left(\frac{n}{\log N}\right)$. I want to show that the running time $T = \Theta\left(\frac{n^2}{\log^2n}\right)$ by first showing that $N = \Theta\left(\frac{n}{\log n}\right)$.

Since $N \leq n$, $N = \Theta\left(\frac{n}{\log N}\right)=\Omega\left(\frac{n}{\log n}\right)$. I am having trouble showing $ N = \Theta\left(\frac{n}{\log N}\right)= O\left(\frac{n}{\log n}\right)$ to be able to use the $\Theta$ notation.

In general, I am not so sure what kind of manipulations are allowed inside $\Theta$ so that its definition is intact.
There is not any strange thing here. As:

$$\frac{n}{\log(n)} = \frac{n}{\log(N\log(N))} = \frac{n}{\log(N)+\log(\log(N))} = \Theta\left(\frac{n}{\log(N)}\right) = O\left(\frac{n}{\log(N)}\right)$$

Therefore, all of them are true.