@Arjun
Sir can you please verify my answer
from observation we can say time complexity of g(n) is O(log n)
hence time complexity for g($2^{n-1}$) will be O(log($2^{n-1}$)) = O(n)
Now
f(n-1) + g(n) + g(pow(2, n-1));
this statement will be called n times
f(n-1) + g(n) + g(pow(2, n-1));
Not doing any work O(log n) O(n)
hence overall max( n times x O(log n) , n times x O(n))
hecne times complexity is O($n^{2}$)