Asymptotic order of a recursive function
What you are asking is really, how many times do you need to apply the transformation $m \mapsto \frac{m}{\log m}$ in order for a value to drop from $n$ below some constant $C$; if the answer is $f(n)$, then $T(n) = \Theta(f(n))$.
Given $n$, define a sequence by $n_0 = n$ and $n_{i+1} = n_i/ \log n_i$. We would like to know the minimal value of $t$ which satisfies $n_t &lt; C$.
Taking logarithm, we see that $\log n_{i+1} = \log n_i - \log\log n_i \geq \log n_i - \log\log n$. Since $\log n_0 = \log n$, this shows that $t \gtrsim \frac{\log n}{\log\log n}$.
In the other direction, as long as $n_i \geq \sqrt{n}$, we have $\log n_{i+1} \leq \log n_i - \log \log \sqrt{n} \leq \log n_i - \log \log n + \log(1/2)$. Hence we get from $n$ down to $\sqrt{n}$ in roughly $\frac{\log n}{\log\log n}$ steps. We then get down to $\sqrt[4]{n}$ in roughly $\frac{\log \sqrt{n}}{\log\log \sqrt{n}} \approx \frac{\frac{1}{2} \log n}{\log\log n}$ steps. Continuing in this way, we see that $t \lesssim \frac{2\log n}{\log\log n}$ (summing the series $1+\frac{1}{2}+\frac{1}{4} + \cdots = 2$).