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I meant: exp(0.5 * log x).

best way :

double a = (double) num;
double x = 1;

// For loop to get the square root value of the entered number.
for( int i = 0; i < n; i++)
{
x = 0.5 * ( x+a / x );
}

System.out.println("sqrt of " + a + " is " + x);

Using binary search actually is the worst way to do that. Because when the square just get to 10 the x = 100 and the time used to find 100's square is log2(100) almost equal to 7(becz 2^7 = 128) just a little bit overruns the linear search. 

My way is to check linearly but a little bit different:

int Find_SR(int x)
{
	vector<int> T;
	int rest = x; /*to store the rest factor*/
	for(int i=2;i<n;)
	{
		if(rest%(i*i)==0)
		{
			rest = rest/(i*i);
			T.push_back(i);
			i=2;
		}
		else
		{
			i++;
		}
	}
	// using for loop to print out the factor that stored in vector.
	
}