用于将log a计算为任意基数b(logb a)的程序

给定两个整数a和b ，任务是找到a到任何底数b的对数，即log _b a 。
例子：

Input: a = 3, b = 2
Output: 1

Input: a = 256, b = 4
Output: 4

使用内置日志函数

1. 借助log()方法找到以2为底的a的日志
2. 借助log()方法找到b的对数以2为底的对数
3. 将计算出的日志a从日志b中除以得到日志_b a ，即$\LARGE log_{b}\text{ } a = \frac{log\text{ }a}{log\text{ }b}$

下面是上述方法的实现

C++

// C++ program to find log(a) on any base b
#include 
using namespace std;
 
int log_a_to_base_b(int a, int b)
{
    return log(a) / log(b);
}
 
// Driver code
int main()
{
    int a = 3;
    int b = 2;
    cout << log_a_to_base_b(a, b) << endl;
 
    a = 256;
    b = 4;
    cout << log_a_to_base_b(a, b) << endl;
 
    return 0;
}
 
// This code is contributed by shubhamsingh10

C

// C program to find log(a) on any base b
 
#include 
#include 
 
int log_a_to_base_b(int a, int b)
{
    return log(a) / log(b);
}
 
// Driver code
int main()
{
    int a = 3;
    int b = 2;
    printf("%d\n",
           log_a_to_base_b(a, b));
 
    a = 256;
    b = 4;
    printf("%d\n",
           log_a_to_base_b(a, b));
 
    return 0;
}

Java

// Java program to find log(a) on any base b
class GFG
{
     
    static int log_a_to_base_b(int a, int b)
    {
        return (int)(Math.log(a) / Math.log(b));
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int a = 3;
        int b = 2;
        System.out.println(log_a_to_base_b(a, b));
     
        a = 256;
        b = 4;
        System.out.println(log_a_to_base_b(a, b));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to find log(a) on any base b
from math import log
 
def log_a_to_base_b(a, b) :
    return log(a) // log(b);
 
# Driver code
if __name__ == "__main__" :
     
    a = 3;
    b = 2;
    print(log_a_to_base_b(a, b));
 
    a = 256;
    b = 4;
    print(log_a_to_base_b(a, b));
 
# This code is contributed by AnkitRai01

C#

// C# program to find log(a) on any base b
 
using System;
 
public class GFG
{
     
    static int log_a_to_base_b(int a, int b)
    {
        return (int)(Math.Log(a) / Math.Log(b));
    }
     
    // Driver code
    public static void Main()
    {
        int a = 3;
        int b = 2;
        Console.WriteLine(log_a_to_base_b(a, b));
     
        a = 256;
        b = 4;
        Console.WriteLine(log_a_to_base_b(a, b));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript program to find log(a) on any base b
 
function log_a_to_base_b(a, b)
{
    return parseInt(Math.log(a) / Math.log(b));
}
 
// Driver code
var a = 3;
var b = 2;
document.write(log_a_to_base_b(a, b) + "<br>");
a = 256;
b = 4;
document.write(log_a_to_base_b(a, b));
 
// This code is contributed by rutvik_56.

输出：

1
4

使用递归

1. 递归地将a除以b，直到a大于b。
2. 计算可能的除法次数。这是a到基数b的对数，即log _b a

下面是上述方法的实现

C++

// C++ program to find log(a) on
// any base b using Recursion
#include 
using namespace std;
 
// Recursive function to compute
// log a to the base b
int log_a_to_base_b(int a, int b)
{
    return (a > b - 1)
            ? 1 + log_a_to_base_b(a / b, b)
            : 0;
}
 
// Driver code
int main()
{
    int a = 3;
    int b = 2;
    cout << log_a_to_base_b(a, b) << endl;
 
    a = 256;
    b = 4;
    cout << log_a_to_base_b(a, b) << endl;
 
    return 0;
}
 
// This code is contributed by shubhamsingh10

C

// C program to find log(a) on
// any base b using Recursion
 
#include 
 
// Recursive function to compute
// log a to the base b
int log_a_to_base_b(int a, int b)
{
    return (a > b - 1)
               ? 1 + log_a_to_base_b(a / b, b)
               : 0;
}
 
// Driver code
int main()
{
    int a = 3;
    int b = 2;
    printf("%d\n",
           log_a_to_base_b(a, b));
 
    a = 256;
    b = 4;
    printf("%d\n",
           log_a_to_base_b(a, b));
 
    return 0;
}

Java

// Java program to find log(a) on
// any base b using Recursion
class GFG
{
     
    // Recursive function to compute
    // log a to the base b
    static int log_a_to_base_b(int a, int b)
    {
        int rslt = (a > b - 1)? 1 + log_a_to_base_b(a / b, b): 0;
        return rslt;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int a = 3;
        int b = 2;
        System.out.println(log_a_to_base_b(a, b));
     
        a = 256;
        b = 4;
        System.out.println(log_a_to_base_b(a, b));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to find log(a) on
# any base b using Recursion
 
# Recursive function to compute
# log a to the base b
def log_a_to_base_b(a, b) :
 
    rslt = (1 + log_a_to_base_b(a // b, b)) if (a > (b - 1)) else 0;
             
    return rslt;
     
# Driver code
if __name__ == "__main__" :
 
    a = 3;
    b = 2;
    print(log_a_to_base_b(a, b));
 
    a = 256;
    b = 4;
    print(log_a_to_base_b(a, b));
 
# This code is contributed by AnkitRai01

C＃

// C# program to find log(a) on
// any base b using Recursion
using System;
 
class GFG
{
     
    // Recursive function to compute
    // log a to the base b
    static int log_a_to_base_b(int a, int b)
    {
        int rslt = (a > b - 1)? 1 + log_a_to_base_b(a / b, b): 0;
        return rslt;
    }
     
    // Driver code
    public static void Main()
    {
        int a = 3;
        int b = 2;
        Console.WriteLine(log_a_to_base_b(a, b));
     
        a = 256;
        b = 4;
        Console.WriteLine(log_a_to_base_b(a, b));
    }
}
 
// This code is contributed by Yash_R

输出：

1
4