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 ?> |
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 = 12print(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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