To be a power of 2 number,is to have a single 1 bit, and the
rest bits are zeros, lik2 1, 2, 4 , 8, 16, 32, 64, 128, ...

the bitsize of the number is sizeof(number) * 8

isPowerOf2() returns 1 if successful, or 0 (false) otherwise
int isPowerOf2 (number)
{
   int foundOnes = 0;
   int bitsize = sizeof(number) * 8;

   for (i = 0; i < bitsize; i++)
   {
      if (number & 1)
      {
        if(++foundOnes > 1)
           return false;
        /* shift the number to the right */
        number >> 1;        
      }
   }
   return foundOnes;
}

int ispow2(int number)
{
 if(n<o) {
 return 0;
 }

 else {
  
 return !(number&(number-1));

 }

int isPowrOf2 (unsigned int number)
{
   float x;

   if (number <= 0) return (0);

   x = log(number) / log(2);
   return ( (int) ( ((int)x*log(2))/log(number)));
}

int isPowerOf2(unsigned int n)
{
  float r=n;
  while(r>1) r/=2.0;
   return (r==1)?1:0; /* 1-> TRUE; 0-> FALSE*/
}

int ispwrof2(int num)
{
	int temp = 0;
	if(num <= 0)
		return 0;
	else if(num > 0)
	{
		while(num%2 == 0)
		{
			temp = num/2;
			num = temp;

		}

	 }
	 if(num == 1)
		return 1;
	 else
		return 0;
}

Answer 2  Ms is correct and more efficient.

because if a number is power 2 it is in form 

1000---->8
100----->4
10------>2  like form
subtract one from that and do with bitwise and it should be
zero.

ie 1000(8) & 0111(7)== 0

#include<stdio.h>
main()
{
int i;
if(i%2=0)
printf("The given number is a power of 2");
else
printf("The given number is not a power of 2");
return 0;
}

#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
 int num,i=0,flag;
 clrscr();
 printf("\nEnter the number:=");
 scanf("%d",&num);
 
 if(num==0 || num==1 || (num%2)!=0) /* Validation Part */
 {
    printf("\n We can't express the given number to power 
of 2");
    return;
  }

 for(;;}
 {
   if(pow(2,i)==num)
   {
     flag=1;
     break;
  }
   if(pow(2,i)>num)
   {
     flag=0;
     break;
   }
i++;
}
if(falg==1)
 printf("\n %d number is 2 power of %d",num,i);
else
 printf("\n%d number can't be expressed as power of 2",num);

getch();
}

#include<stdio.h>
#include<conio.h>

int main()
{
	int i,count = 0,number;

	printf("Enter the number\n");
	scanf("%d",&number);

	for(i=15;i>=0;i--)
	{
		if((1<<i) & number)
			count++;
	}

	if(count == 1)
		printf("\nThe Given Number is Power of 2\n");
	else
		printf("\nThe Given Number is Not Power of 2\n");

}

unsigned int is_power_of_2(unsigned int x)
{
return (x != 0) && ((x & (x - 1)) == 0);
}

#include<stdio.h>
int main()
{
int num,i=1;
printf("Enter the number:");
scanf("%d", &num);
do
{
i=i*2;
if(i==num)
{
printf("\n The no.is power of 2.");
break;
}
else if(i>num)
{
printf("\n The no. is not power of 2");
break;
}
}
while(i != num);
return 0;
}

/* THIS SOLUTION IS 100% CORRECT */

#include<stdio.h>
int isPowerOf2(float n)
{
  while(n>1)
  {
  n/=2.0;
  }
  return (n==1)?1:0; /* 1-> TRUE; 0-> FALSE*/
}
void main()
{
  int x,n;
  printf("enter a number\n");
  fflush(stdin);
  scanf("%d",&n);
  x=isPowerOf2(n);
  if(x==1)
  printf("\n yes");
  else
  printf("\n no");
}