Square root of a number using log
Last Updated :
25 Sep, 2022
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For a given number find the square root using log function. Number may be int, float or double.
Examples:
Input : n = 9
Output : 3Input : n = 2.93
Output : 1.711724
We can find square root of a number using sqrt() method:
// 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 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 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# 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 program to demonstrate finding
// square root of a number using sqrt()
$n = 12;
echo sqrt($n);
// This code is contributed by jit_t
?>
<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
Time complexity: O(log2n), for using sqrt() function.
Auxiliary space: O(1)
We can also find square root using log2() library function:
// 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 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 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# 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 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
?>
<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
3.464102
Time complexity: O(log2log2N), complexity of using log(N) is log(logN), and pow(x,N) is log(N), so pow(2,0.5*log(n)) will be log(logN).
Auxiliary space: O(1)
How does the 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))
