This is done by using exponential theory :-

 I am giving my code here for a^b


int power (int a,int b)
{
   int x=1,y=a;
   while(b>0)
   {
       if(b%2==1)  x=(x*y)
       if (b/=2) y=(y*y);
   }
 return x;
}


Run time  O(log b)

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