Another way of thinking about this question is calling log₂(n) of k. With that we have n=2^k. When n grows asymptotically (approach infinity), k also grows.

So 5^log₂(n) (call it F(N) like Sumeet) becomes 5^k and n^1/2 (call it G(N) as well) becomes 2^k/2.

We can clearly see that G(N) = O(F(N)), since 2^k/2 <= 5^k for k >= 1.

Now we just need to show that G(N) = Omega(F(N)) to prove G(N) = Theta(F(N)).

To do so, we need to find a positive constant c such as:

c . F(N) <= G(N) for n > n₀, that is c <= G(N) / F(N) = 2^k/2 / 5^k. At first we think that this ratio will go to zero when k approaches infinity, but look:

We can make 2^k/2= 2^(k.m)/(2.m)=(2^m/2)^k/m. We can choose m such as 2^m/2=5. To do so we need m>4 (keep this inequality in mind, it is important). With this trick we can transform 2^k/2 into 5^k/m so or ratio becomes 5^k/m / 5^k that equals 5^k.(1-m)/m. Since m < 4 we have 1 - m < 0, so the exponent of the previous exponential is negative:

5^k(1-m)/m = 1/(5^k.(m-1).m) = 1/(5^b) where b = k.(m-1).m.

Since m is a constant greater than 4, when k goes to infinity (equivalent to n goes to infinity) the ratio 2^k/2 / 5^k that is equivalent to 1/(5^b) with b going to infinity also goes to infinity!

So if we pick a value of n₀ really high (asymptotically high) we can pick any positive constant c to make c . F(N) <= G(N). That is G(N) = O(F(N)) AND G(N) = Omega(F(N)), so G(N) = Theta(F(N)).