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.

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++ 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 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 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# 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 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
?>

<script>
// javascript code for finding greater
// between the a^b and b^a

    // Function to find the greater value
    function findGreater(a , b) {
        var x =  a * (Math.log( (b)));
        var y =  b * (Math.log( (a)));
        if (y > x) {
            document.write("a^b is greater<br/>");
        } else if (y < x) {
            document.write("b^a is greater<br/>");
        } else {
            document.write("Both are equal<br/>");
        }
    }

    // Driver code
    
        var a = 3, b = 5, c = 2, d = 4;
        findGreater(a, b);
        findGreater(c, d);

// This code is contributed by todaysgaurav 
</script>

Output: 
 

a^b is greater
Both are equal

Time Complexity: O(logn), since it is using inbuilt log function

Auxiliary Space: O(1), since no extra space has been taken.