How to solve the recurrence relation:
$\displaystyle T(n) = \begin{cases} \sqrt{2T\left(\frac n2\right)} + \log_2(n) & \mbox{if }\ n>2\\ 1 & \mbox{if }\ n\leq2 \end{cases}$
n is set of all real numbers
I am not sure how to remove the sqrt, by some substitution. i tried multiple methods by replacing with a different function but i was unable to remove it.
Judging by the name of your function and the base $2$ logarithm, this recurrence arises from estimating the running time of an algorithm, in which case we only care about the value of $T(n)$ up to a multiplicative factor as $n\to\infty$. We may apply a variant of the master theorem to resolve such questions.
In this case, it is convenient to change variables by setting $a_k=T(2^k)$ so that $a_{k}=\sqrt{2a_{k-1}}+k$. Clearly $a_k>k$, and a simple induction shows that $a_k<2k$ as well. Therefore $T(n)$ grows logarithmically. (In fact, with more work we can get that $T(n)=(1+o(1))\log_2(n)$.