Recursive Algorithm Analysis $$T(n) = 2\cdot \sqrt{n} \cdot T(\sqrt{n}) + \Theta (\lg n)$$

I have been trying to solve this question but I could not find anything.

My approach:

$n = 2^k$  

$S(k) = T(2^n)$ and $S(k/2) = T(2^{n/2})$

Finally:
$S(k) = 2^{1+k/2} \cdot S(k/2) + c \cdot \lg(k)  $

After that, I tried to build recursion tree but I can not find the sum. Do you have any ideas?

Thanks in advance.
Let's consider the recurrence without the big $\Theta$. Then (using base two logarithms)
$$
\begin{align*}
T(n) &amp;= 2n^{1-1/2} T(n^{1/2}) + \log n \\ &amp;=
4n^{1-1/4} T(n^{1/4}) + \log n (1 + n^{1-1/2}) \\ &amp;=
8n^{1-1/8} T(n^{1/8}) + \log n (1 + n^{1-1/2} + n^{1-1/4}) \\ &amp;= \cdots \\ &amp;=
(\log n) n^{1-1/\log n} T(n^{1/\log n}) + \log n (1 + n^{1-1/2} + n^{1-1/4} + \dots + n^{1-1/\log n}) \\ &amp;=
2n\log n T(2) + n\log n \left(\frac{1}{n} + \frac{1}{n^{1/2}} + \dots + \frac{1}{2}\right) \\ &amp;= \Theta(n\log n).
\end{align*}
$$