Larger of a^b or b^a (a raised to power b or b raised to power a)

Given two numbers $a and b$ , find which is greater $a^b or \, b^a$ .

If $a^b > b^a$ , print a^b is greater
If $a^b < b^a$ , print b^a is greater
If $a^b = b^a$ , print Both are equal

Examples:

Input : 3 5
Output : a^b is greater
3^5 = 243, 5^3 = 125. Since, 243>125, therefore a^b > b^a.

Input : 2 4
Output : Both are equal
2^4 = 16, 4^2 = 16. Since, 16=16, therefore a^b = b^a.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute Force solution would be to just compute $a^b or \, b^a$ and compare them. But since $a and b$ can be large enough that $a^b or \, b^a$ can not be stored even in long long int, so this solution is not feasible. Also computing to the power n would require at least $O(logn)$ time using the fast exponentiation technique.

Efficient approach would be to use logarithm. We have to compare $a^b or \, b^a$ . If we take log, the problem reduces to comparing $\log_a b \, and \, \log_b a$ .
Hence,

If $b\log a > a\log b$ , print a^b is greater
If $b\log a < a\log b$ , print b^a is greater
If $b\log a = a\log b$ , print Both are equal

Below is the implementation of the efficient approach discussed above.

C++

// C++ code for finding greater 
// between the a^b and b^a 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the greater value 
void findGreater(int a, int b) 
{ 
    long double x = (long double)a * (long double)(log((long double)(b))); 
    long double y = (long double)b * (long double)(log((long double)(a))); 
    if (y > x) { 
        cout << "a^b is greater" << endl; 
    } 
    else if (y < x) { 
        cout << "b^a is greater" << endl; 
    } 
    else { 
        cout << "Both are equal" << endl; 
    } 
} 
  
// Driver code 
int main() 
{ 
    int a = 3, b = 5, c = 2, d = 4; 
    findGreater(a, b); 
    findGreater(c, d); 
    return 0; 
} 

Java

// Java code for finding greater 
// between the a^b and b^a 
  
public class GFG{ 
  
    // Function to find the greater value 
    static void findGreater(int a, int b) 
    { 
        double x = (double)a * (double)(Math.log((double)(b))); 
        double y = (double)b * (double)(Math.log((double)(a))); 
        if (y > x) { 
            System.out.println("a^b is greater") ; 
        } 
        else if (y < x) { 
            System.out.println("b^a is greater") ; 
        } 
        else { 
            System.out.println("Both are equal") ; 
        } 
    } 
      
    // Driver code 
    public static void main(String []args) 
    { 
        int a = 3, b = 5, c = 2, d = 4; 
        findGreater(a, b); 
        findGreater(c, d); 
    } 
    // This code is contributed by Ryuga 
} 

Python 3

# Python 3 code for finding greater 
# between the a^b and b^a 
import math 
  
# Function to find the greater value 
def findGreater(a, b): 
  
    x = a * (math.log(b)); 
    y = b * (math.log(a)); 
    if (y > x): 
        print ("a^b is greater"); 
    elif (y < x):  
        print("b^a is greater"); 
    else : 
        print("Both are equal");  
  
# Driver code 
a = 3; 
b = 5; 
c = 2; 
d = 4; 
findGreater(a, b); 
findGreater(c, d); 
  
# This code is contributed  
# by Shivi_Aggarwal 

C#

// C# code for finding greater 
// between the a^b and b^a 
   
using System; 
public class GFG{ 
   
    // Function to find the greater value 
    static void findGreater(int a, int b) 
    { 
        double x = (double)a * (double)(Math.Log((double)(b))); 
        double y = (double)b * (double)(Math.Log((double)(a))); 
        if (y > x) { 
            Console.Write("a^b is greater\n") ; 
        } 
        else if (y < x) { 
            Console.Write("b^a is greater"+"\n") ; 
        } 
        else { 
            Console.Write("Both are equal") ; 
        } 
    } 
       
    // Driver code 
    public static void Main() 
    { 
        int a = 3, b = 5, c = 2, d = 4; 
        findGreater(a, b); 
        findGreater(c, d); 
    } 
      
} 

PHP

<?php 
// PHP code for finding greater 
// between the a^b and b^a 
  
// Function to find the greater value 
function findGreater($a, $b) 
{ 
    $x = (double)$a * (double)(log((double)($b))); 
    $y = (double)$b * (double)(log((double)($a))); 
    if ($y > $x)  
    { 
        echo "a^b is greater", "\n"; 
    } 
    else if ($y < $x)  
    { 
        echo "b^a is greater", "\n" ; 
    } 
    else 
    { 
        echo "Both are equal", "\n" ; 
    } 
} 
  
// Driver code 
$a = 3; 
$b = 5; 
$c = 2; 
$d = 4; 
findGreater($a, $b); 
findGreater($c, $d); 
  
// This code is contributed by ajit 
?> 

Output:

a^b is greater
Both are equal

Time Complexity: $O(1)$

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

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