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

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
?>

Javascript

<script>

// Javascript program to demonstrate finding
// square root of a number using sqrt()
var n = 12;

document.write(Math.sqrt(n).toFixed(6));

// This code is contributed by aashish1995

</script>

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
?>

Javascript

<script>
    // Javascript program to demonstrate finding
    // square root of a number using log2()

    function squareRoot(n)
    {
         return Math.pow(2, 0.5 * (Math.log(n) /Math.log(2)));
    }

    let n = 12;
      document.write(squareRoot(n).toFixed(15));

</script>

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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