Logarithm - GeeksforGeeks

Logarithmic function is the inverse to the exponential function. A logarithm to the base b is the power to which b must be raised to produce a given number. For example, $\log_2 8$ is equal to the power to which 2 must be raised to in order to produce 8. Clearly, 2^3 = 8 so $\log_2 8$ = 3. In general, for b > 0 and b not equal to 1.

Fact about Logarithm :

Logarithms were quickly adopted by scientists because of various useful properties that simplified long, tedious calculations.
Logarithm to base 10 (that is b = 10) is called the common logarithm and has many applications in science and engineering.
Natural logarithm, is a logarithm with base e. It is used in mathematics and physics, because of its simpler derivative.
Binary logarithm is a logarithm with base 2 and is commonly used in computer science.

Laws of Logarithms :

Laws	Description
$\log_a bc$	$\log_a b$ + $\log_a c$
$\log_a b/c$	$\log_a b$ – $\log_a b$
$\log_a b^c$	$c\log_a b$
$\log_a 1/b$	$-\log_a b$
$\log_a 1$	$0$
$\log_a a$	$1$
$\log_a a^r$	$r$
*$\log_a b$ $\log_b c$**	$\log_a c$
$\log_b a$	$1/\log_a b$

How to find logarithm of number?

Naive solution:
The idea is to create a function that calculates and returns $\log_2 n$ . For example, if n = 64, then your function should return 6, and if n = 129, then your function should return 7.

C

// C program to find log(n) using Recursion 
#include <stdio.h> 
  
unsigned int Log2n(unsigned int n) 
{ 
    return (n > 1) ? 1 + Log2n(n / 2) : 0; 
} 
  
int main() 
{ 
    unsigned int n = 32; 
    printf("%u", Log2n(n)); 
    getchar(); 
    return 0; 
} 

Java

// Java program to find log(n) 
// using Recursion 
class Gfg1 { 
  
    static int Log2n(int n) 
    { 
        return (n > 1) ? 1 + Log2n(n / 2) : 0; 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        int n = 32; 
        System.out.println(Log2n(n)); 
    } 
} 

Python3

# Python 3 program to 
# find log(n) using Recursion 
  
def Log2n(n): 
  
    return 1 + Log2n(n / 2) if (n > 1) else 0
  
# Driver code 
n = 32
print(Log2n(n)) 

C#

// C# program to find log(n) 
// using Recursion 
using System; 
  
class GFG { 
  
    static int Log2n(int n) 
    { 
        return (n > 1) ? 1 + Log2n(n / 2) : 0; 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int n = 32; 
  
        Console.Write(Log2n(n)); 
    } 
} 

PHP

<?php 
// PHP program  
// to find log(n) using Recursion 
  
function Log2n($n) 
{ 
    return ($n > 1) ? 1 + Log2n($n / 2) : 0; 
} 
  
// Drive main 
  
    $n = 32; 
    echo Log2n($n); 
?> 