$T(n)=2T(n^{1/2})+log (n)$
$T(n^{1/2})=2T(n^{1/4})+log (n^{1/2})$
$T(n^{1/4})=2T(n^{1/8})+log (n^{1/4})$
Now Back-Substituting:
$T(n)=2T(n^{1/2})+log (n)$
$T(n)=2(2T(n^{1/4})+log (n^{1/2}))+log (n)$
$T(n)=2^2T(n^{1/4})+2log (n^{1/2})+log (n)$
$T(n)=2^2(2T(n^{1/8})+log (n^{1/4}))+2log (n^{1/2})+log (n)$
$T(n)=2^3T(n^{1/8})+2^2log (n^{1/4})+2log (n^{1/2})+log (n)$
And it can be written as:
$T(n)=2^3T(n^{1/8})+3log(n)$
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$T(n)=2^kT(n^{1/2^k})+klog(n)$
Now , for sake of simplicity I’ll consider T(2) = 1
$n^{1/2^k} = 2$
Taking log both sides
$log (n^{1/2^k}) = log 2$
$1/2k\hspace{1mm}log(n) = 1$
$log(n) = 2^k$
$k = loglog(n)$
Now substituting this into our answer :
$T(n)=2^{loglog (n)}+loglog(n) \hspace{1mm} log(n)$
$T(n)=log (n)+loglog(n) \hspace{1mm} log(n)$
And,
$log (n) < loglog(n) \hspace{1mm} log(n)$
So,
$T(n)=O(loglog(n) \hspace{1mm} log(n))$
Answer : B