Square root of a number using log

For a given number find the square root using log function. Number may be int, float or double.

Examples:

Input  : n = 9
Output : 3

Input  : n = 2.93
Output : 1.711724

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can find square root of a number using sqrt() method.

C++

// C++ program to demonstrate finding 
// square root of a number using sqrt() 
#include<bits/stdc++.h> 
  
int main(void) 
{ 
    double n = 12; 
    printf("%lf ", sqrt(n)); 
    return 0; 
} 

Java

// Java program to demonstrate finding 
// square root of a number using sqrt() 
  
import java.io.*; 
  
class GFG { 
    public static void main (String[] args) { 
    double n = 12; 
    System.out.println(Math.sqrt(n)); 
  
  
// This code is contributed by akt_mit 
    } 
} 

Python3

# Python3 program to demonstrate finding 
# square root of a number using sqrt() 
import math 
  
if __name__=='__main__': 
    n = 12
    print(math.sqrt(n)) 
  
# This code is contributed by 
# Sanjit_Prasad 

C#

// C# program to demonstrate finding 
// square root of a number using sqrt() 
using System; 
  
class GFG 
{ 
public static void Main() 
{ 
    double n = 12; 
    Console.Write(Math.Sqrt(n)); 
} 
} 
  
// This code is contributed 
// by Akanksha Rai 

PHP

<?php 
// PHP program to demonstrate finding  
// square root of a number using sqrt()  
$n = 12;  
echo sqrt($n);  
  
// This code is contributed by jit_t 
?> 

Output :

3.464102

We can also find square root using log2() library function:

C++

// C++ program to demonstrate finding 
// square root of a number using log2() 
#include<bits/stdc++.h> 
  
double squareRoot(double n) 
{ 
    return pow(2, 0.5*log2(n)); 
} 
  
int main(void) 
{ 
    double n = 12; 
    printf("%lf ", squareRoot(n)); 
    return 0; 
} 

Java

// Java program to demonstrate finding  
// square root of a number using log2()  
import java.io.*; 
  
class GFG  
{ 
static double squareRoot(double n)  
{  
    return Math.pow(2, 0.5 * (Math.log(n) / 
                              Math.log(2)));  
}  
  
// Driver Code 
public static void main (String[] args)  
{ 
    double n = 12;  
    System.out.println(squareRoot(n));  
} 
} 
  
// This code is contributed by akt_mit  

Python

# Python program to demonstrate finding 
# square root of a number using sqrt() 
import math 
  
# function to return squareroot 
def squareRoot(n): 
  
    return pow(2, 0.5 * math.log2(n)) 
  
# Driver program 
  
n = 12
print(squareRoot(n)) 
  
# This code is contributed by 
# Sanjit_Prasad 

C#

// C# program to demonstrate finding  
// square root of a number using log2() 
using System; 
  
public class GFG{ 
      
static double squareRoot(double n)  
{  
     return Math.Pow(2, 0.5 * (Math.Log(n) /Math.Log(2))); 
}  
  
      
    static public void Main (){ 
            double n = 12;  
            Console.WriteLine(squareRoot(n));  
    } 
//This code is contributed by akt_mit     
} 

PHP

<?php 
// PHP program to demonstrate finding  
// square root of a number using log2()  
function squareRoot($n)  
{  
    return pow(2, 0.5 * log($n, 2));  
}  
  
// Driver Code 
$n = 12;  
echo squareRoot($n);  
      
// This code is contributed by ajit 
?> 

Output:

3.464101615137755

How does above program work?

 let d be our answer for input number n
 then n^(1/2) = d 
     apply log2 on both sides
      log2(n^(1/2)) = log2(d)
      log2(d) = 1/2 * log2(n)
      d = 2^{(1/2 * log2(n))}
      d = pow(2, 0.5*log2(n))

This article is contributed by Tumma Umamaheswararao from Jntuh College of Engineering . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python

C#

PHP

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