Check if a number can be expressed as power | Set 2 (Using Log)

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Check if a number can be expressed as x^y (x raised to power y)
Given a positive integer n, find if it can be expressed as x^y where y > 1 and x > 0. x and y both are integers.
Examples :

Input:  n = 8
Output: true
8 can be expressed as 2^3

Input:  n = 49
Output: true
49 can be expressed as 7^2

Input:  n = 48
Output: false
48 can't be expressed as x^y

We have discussed two different approaches in below post.
Check if a number can be expressed as x^y (x raised to power y).
The idea is find Log n in different bases from 2 to square root of n. If Log n for a base becomes integer then result is true, else result is false.

C++

// CPP program to find if a number
// can be expressed as x raised to
// power y.
#include <bits/stdc++.h>
using namespace std;
 
bool isPower(unsigned int n)
{
    // Find Log n in different bases
    // and check if the value is an
    // integer
    for (int x=2; x<=sqrt(n); x++) {
        float f = log(n) / log(x);
        if ((f - (int)f) == 0.0) 
            return true;        
    }
    return false;
}
 
// Driver code
int main()
{
    for (int i = 2; i < 100; i++)
        if (isPower(i))
            cout << i << "  ";
    return 0;
}

Java

// Java program to find if a number
// can be expressed as x raised to
// power y.
class GFG {
     
    static boolean isPower(int n)
    {
        // Find Log n in different
        // bases and check if the 
        // value is an integer
        for (int x = 2; x <= 
               (int)Math.sqrt(n); x++) 
        {
            float f = (float)Math.log(n) / 
                      (float) Math.log(x);
                       
            if ((f - (int)f) == 0.0) 
                return true;     
        }
        return false;
    }
     
    // Driver code 
    public static void main(String args[])
    {
        for (int i = 2; i < 100; i++)
            if (isPower(i))
                System.out.print( i + " ");
    }
}
 
// This code is contributed by Sam007

Python3

# Python3 program to find if a number
# can be expressed as x raised to
# power y.
import math
 
def isPower(n):
 
    # Find Log n in different 
    # bases and check if the 
    # value is an integer
    for x in range(2,int(math.sqrt(n)) + 1): 
     
        f = math.log(n) / math.log(x);
        if ((f - int(f)) == 0.0): 
            return True;     
     
    return False;
 
# Driver code
for i in range(2, 100):
    if (isPower(i)):
        print(i, end = " ");
 
# This code is contributed by mits

C#

// C# program to find if a number
// can be expressed as x raised to
// power y.
using System;
 
class GFG 
{
    static bool isPower(int n)
    {
        // Find Log n in different
        // bases and check if the 
        // value is an integer
        for (int x = 2; 
                 x <= (int)Math.Sqrt(n); x++) 
        {
            float f = (float)Math.Log(n) / 
                      (float) Math.Log(x);
            if ((f - (int)f) == 0.0) 
                return true;     
        }
        return false;
    }
    // Driver Code
    public static void Main()
    {
        for (int i = 2; i < 100; i++)
            if (isPower(i))
                Console.Write( i + " ");
    } 
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP program to find if a number
// can be expressed as x raised to
// power y.
 
function isPower($n)
{
    // Find Log n in different 
    // bases and check if the 
    // value is an integer
    for ($x = 2; $x <= sqrt($n); $x++) 
    {
        $f = log($n) / log($x);
        if (($f - (int)$f) == 0.0) 
            return true;     
    }
    return false;
}
 
// Driver code
for ($i = 2; $i < 100; $i++)
    if (isPower((int)$i))
        echo $i." ";
 
// This code is contributed by Sam007
?>

Javascript

<script>
 
// javascript program to find if a number
// can be expressed as x raised to
// power y. 
function isPower(n) {
        // Find Log n in different
        // bases and check if the
        // value is an integer
        for (x = 2; x <= parseInt( Math.sqrt(n)); x++) 
        {
            var f =  Math.log(n) /  Math.log(x);
 
            if ((f - parseInt( f)) == 0.0)
                return true;
        }
        return false;
    }
 
    // Driver code
     
        for (i = 2; i < 100; i++)
            if (isPower(i))
                document.write(i + " ");
 
// This code contributed by Rajput-Ji
 
</script>

Output:

4  8  9  16  25  27  32  36  49  64  81

Time Complexity : O(sqrt(N))

Auxiliary Space : O(1) ,as we are not using any extra space

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