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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
    }
}

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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))  

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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