Check if a number is a power of another

Last Updated : 15 Apr, 2026

Given two positive numbers x and y, check if y is a power of x or not.

Examples : 

Input:  x = 2, y = 8
Output: true
Explanation: 23 is equal to 8

Input:  x = 10, y = 1
Output: true
Explanation: 100 is equal to 1

Input:  x = 10, y = 1001
Output: false

Try It Yourself
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Table of Content

[Naive Approach] Repeated Multiplication Method

The idea is to start from 1 and repeatedly multiplying it by x. If the value becomes equal to y, then y is a power of x and if it exceeds y without matching, then it is not. This method relies on simple multiplication and avoids using logarithms or recursion.

Dry run for x = 2 and y = 8:

  1. Start with pow = 1.
  2. Multiply by 2, now pow = 2, which is still less than 8, so continue.
  3. Multiply again, now pow = 4, still less than 8, so continue.
  4. Multiply again, now pow = 8, which becomes equal to y.
  5. Since pow equals y, the function returns true.

Final conclusion: 8 is a power of 2.

C++ 
#include <iostream>
using namespace std;

bool isPower(int x, int y)
{
    // The only power of 1 is 1 itself
    if (x == 1)
        return (y == 1);

    // Repeatedly compute power of x
    int pow = 1;
    while (pow < y)
        pow *= x;

    // Check if power of x becomes y
    return (pow == y);
}


int main()
{
    cout << boolalpha;  
    cout << isPower(10, 1) << endl;    
    cout << isPower(1, 20) << endl;    
    cout << isPower(2, 128) << endl;   
    cout << isPower(2, 30) << endl;    
    return 0;
}
C 
#include <stdio.h>
#include <stdbool.h>

bool isPower(int x, int y)
{
    // The only power of 1 is 1 itself
    if (x == 1)
        return (y == 1);

    // Repeatedly compute power of x
    int pow = 1;
    while (pow < y)
        pow *= x;

    // Check if power of x becomes y
    return (pow == y);
}

int main()
{
    printf("%s\n", isPower(10, 1) ? "true" : "false");
    printf("%s\n", isPower(1, 20) ? "true" : "false");
    printf("%s\n", isPower(2, 128) ? "true" : "false");
    printf("%s\n", isPower(2, 30) ? "true" : "false");
    return 0;
}
Java 
class GfG {
    
    public static boolean isPower(int x, int y) {
        
        // The only power of 1 is 1 itself
        if (x == 1)
            return (y == 1);

        // Repeatedly compute power of x
        long pow = 1;
        while (pow < y)
            pow *= x;

        // Check if power of x becomes y
        return (pow == y);
    }

    public static void main(String[] args) {
        System.out.println(isPower(10, 1));
        System.out.println(isPower(1, 20));
        System.out.println(isPower(2, 128));
        System.out.println(isPower(2, 30));
    }
}
Python 
def isPower(x, y):
    
    # The only power of 1 is 1 itself
    if x == 1:
        return y == 1

    # Repeatedly compute power of x
    pow = 1
    while pow < y:
        pow *= x

    # Check if power of x becomes y
    return pow == y

if __name__ == '__main__':
    print("true" if isPower(10, 1) else "false")
    print("true" if isPower(1, 20) else "false")
    print("true" if isPower(2, 128) else "false")
    print("true" if isPower(2, 30) else "false")
C# 
using System;

class GfG {
    static bool isPower(int x, long y) {
        // The only power of 1 is 1 itself
        if (x == 1)
            return (y == 1);

        // Repeatedly compute power of x
        long pow = 1;
        while (pow < y)
            pow *= x;

        // Check if power of x becomes y
        return (pow == y);
    }

    static void Main() {
        Console.WriteLine(isPower(10, 1) ? "true" : "false");
        Console.WriteLine(isPower(1, 20) ? "true" : "false");
        Console.WriteLine(isPower(2, 128) ? "true" : "false");
        Console.WriteLine(isPower(2, 30) ? "true" : "false");
    }
}
JavaScript 
function isPower(x, y) {
    // The only power of 1 is 1 itself
    if (x === 1)
        return y === 1;

    // Repeatedly compute power of x
    let pow = 1;
    while (pow < y)
        pow *= x;

    // Check if power of x becomes y
    return pow === y;
}

// Driver code
console.log(isPower(10, 1));
console.log(isPower(1, 20));
console.log(isPower(2, 128));
console.log(isPower(2, 30));

Output
true
false
true
false

Time complexity: O(Logxy), The loop multiplies by x each time, reaching y in logarithmic steps.
Auxiliary space: O(1)

[Better Approach] Exponentiation and Binary Search Method

The idea is to increase the value very quickly using repeated squaring instead of normal multiplication, allowing it to reach close to y in fewer steps. If the value becomes equal to y, it returns true. If it goes beyond y, a binary search is used to check different possible powers of x and find if any of them exactly equals y.

Dry run for x = 2 and y = 8:

  1. Start with pow = 2 and i = 1.
  2. Square pow, now pow = 4 and i = 2, still less than 8 so continue.
  3. Square again, now pow = 16 and i = 4, this exceeds 8 so stop.
  4. Since pow is not equal to 8, perform binary search.
  5. Set range from low = 1 to high = 4 (possible powers).
  6. Find mid = 2, compute 22 = 4, which is less than 8, so move right.
  7. Find mid = 3, compute 23 = 8, which matches y.
  8. Since a matching power is found, the function returns true.

Final conclusion: 8 is a power of 2.

C++ 
#include <iostream>
#include <cmath>
using namespace std;

bool isPower(int x, int y) {
    
    // The only power of 1 is 1 itself
    if (x == 1)
        return (y == 1);

    // If y is 1, then x^0 = 1
    if (y == 1)
        return true;

    // Repeatedly compute power of x using squaring
    int pow = x, i = 1;
    while (pow < y) {
        pow *= pow;
        
        // track exponent growth
        i *= 2;
    }

    if (pow == y)
        return true;

    // Apply binary search to find correct power
    int low = x, high = pow;
    while (low <= high) {
        int mid = low + (high - low) / 2;

        // Estimate exponent using logarithm
        int exponent = (int)(log(mid) / log(x));
        int result = (int)powl(x, exponent);

        if (result == y)
            return true;

        // Move search space
        if (result < y)
            low = mid + 1;
        else
            high = mid - 1;
    }

    // If no power matches
    return false;
}

int main() {
    
    cout << boolalpha;
    cout << isPower(10, 1) << endl;
    cout << isPower(1, 20) << endl;
    cout << isPower(2, 128) << endl;
    cout << isPower(2, 30) << endl;
    return 0;
}
Java 
class GfG {
    
    public static boolean isPower(int x, int y) {
        
        // The only power of 1 is 1 itself
        if (x == 1)
            return (y == 1);

        // If y is 1, then x^0 = 1
        if (y == 1)
            return true;

        // Repeatedly compute power of x using squaring
        int pow = x, i = 1;
        while (pow < y) {
            pow *= pow;
            
            // track exponent growth
            i *= 2;
        }

        if (pow == y)
            return true;

        // Apply binary search to find correct power
        int low = x, high = pow;
        while (low <= high) {
            int mid = low + (high - low) / 2;

            // Estimate exponent using logarithm
            int exponent = (int)(Math.log(mid) / Math.log(x));
            int result = (int)Math.pow(x, exponent);

            if (result == y)
                return true;

            // Move search space
            if (result < y)
                low = mid + 1;
            else
                high = mid - 1;
        }

        // If no power matches
        return false;
    }

    public static void main(String[] args) {
        System.out.println(isPower(10, 1) ? "true" : "false");
        System.out.println(isPower(1, 20) ? "true" : "false");
        System.out.println(isPower(2, 128) ? "true" : "false");
        System.out.println(isPower(2, 30) ? "true" : "false");
    }
}
Python 
import math

def isPower(x, y):
    
    # The only power of 1 is 1 itself
    if x == 1:
        return y == 1
    
    # If y is 1, then x^0 = 1
    if y == 1:
        return True

    # Repeatedly compute power of x using squaring
    pow = x
    i = 1
    while pow < y:
        pow *= pow
        
        # track exponent growth
        i *= 2

    if pow == y:
        return True

    # Apply binary search to find correct power
    low, high = x, pow
    while low <= high:
        mid = low + (high - low) // 2

        # Estimate exponent using logarithm
        exponent = int(math.log(mid) / math.log(x))
        result = int(x ** exponent)

        if result == y:
            return True

        # Move search space
        if result < y:
            low = mid + 1
        else:
            high = mid - 1

    # If no power matches
    return False

if __name__ == "__main__":
    print("true" if isPower(10, 1) else "false")
    print("true" if isPower(1, 20) else "false")
    print("true" if isPower(2, 128) else "false")
    print("true" if isPower(2, 30) else "false")
C# 
using System;

class GfG {
    public static bool isPower(int x, int y) {
        
        // The only power of 1 is 1 itself
        if (x == 1)
            return (y == 1);

        // If y is 1, then x^0 = 1
        if (y == 1)
            return true;

        // Repeatedly compute power of x using squaring
        int pow = x, i = 1;
        while (pow < y) {
            pow *= pow;
            
            // track exponent growth
            i *= 2;
        }

        if (pow == y)
            return true;

        // Apply binary search to find correct power
        int low = x, high = pow;
        while (low <= high) {
            int mid = low + (high - low) / 2;

            // Estimate exponent using logarithm
            int exponent = (int)(Math.Log(mid) / Math.Log(x));
            int result = (int)Math.Pow(x, exponent);

            if (result == y)
                return true;

            // Move search space
            if (result < y)
                low = mid + 1;
            else
                high = mid - 1;
        }

        // If no power matches
        return false;
    }

    static void Main() {
        Console.WriteLine(isPower(10, 1) ? "true" : "false");
        Console.WriteLine(isPower(1, 20) ? "true" : "false");
        Console.WriteLine(isPower(2, 128) ? "true" : "false");
        Console.WriteLine(isPower(2, 30) ? "true" : "false");
    }
}
JavaScript 
function isPower(x, y) {
    
    // The only power of 1 is 1 itself
    if (x === 1)
        return (y === 1);

    // If y is 1, then x^0 = 1
    if (y === 1)
        return true;

    // Repeatedly compute power of x using squaring
    let pow = x, i = 1;
    while (pow < y) {
        pow *= pow;
        
        // track exponent growth
        i *= 2;
    }

    if (pow === y)
        return true;

    // Apply binary search to find correct power
    let low = x, high = pow;
    while (low <= high) {
        let mid = Math.floor(low + (high - low) / 2);

        // Estimate exponent using logarithm
        let exponent = Math.floor(Math.log(mid) / Math.log(x));
        let result = Math.pow(x, exponent);

        if (result === y)
            return true;

        // Move search space
        if (result < y)
            low = mid + 1;
        else
            high = mid - 1;
    }

    // If no power matches
    return false;
}

// Driver code
console.log(isPower(10, 1) ? "true" : "false");
console.log(isPower(1, 20) ? "true" : "false");
console.log(isPower(2, 128) ? "true" : "false");
console.log(isPower(2, 30) ? "true" : "false");

Output
true
false
true
false

Time Complexity: O(log y), First loop grows very fast (squaring) which takes log log y steps, then binary search takes log y, so overall ≈ log y.
Auxiliary Space: O(1)

[Expected Approach] Logarithmic Method - O(1) Time and O(1) Space

By applying the logarithm change of base formula, it computes log⁡x(y)=log⁡(y)/log⁡(x). If the result is an integer, it means that y is an exact power of x, and the function returns true. Otherwise, it returns false.

Dry run for x = 2 and y = 8:

  • Compute log(8) / log(2) using the logarithmic formula.
  • The result comes out to be 3.
  • This value represents the power such that 23 = 8.
  • Now check whether this result is an integer.
  • Since 3 is a whole number, it confirms that 8 is an exact power of 2.
  • Therefore, the function returns true.

Final conclusion: 8 is a power of 2.

C++ 
#include <iostream>
#include <cmath>
using namespace std;

bool isPower(int x, int y)
{
    // Edge case: 1^k = 1 only
    if (x == 1)
        return y == 1;

    // Edge case: x^0 = 1
    if (y == 1)
        return true;

    // Compute logarithm
    double res = log(y) / log(x);
    
    // Compare with rounded value using a small 
    // tolerance to avoid floating point errors
    return fabs(res - round(res)) < 1e-10;
}

int main()
{

    cout << boolalpha;
    cout << isPower(10, 1) << endl;
    cout << isPower(1, 20) << endl;
    cout << isPower(2, 128) << endl;
    cout << isPower(2, 30) << endl;

    return 0;
}
C 
#include <math.h>
#include <stdbool.h>
#include <stdio.h>

bool isPower(int x, int y)
{
    // Edge case: 1^k = 1 only
    if (x == 1)
        return y == 1;

    // Edge case: x^0 = 1
    if (y == 1)
        return true;

    // Compute logarithm
    double res = log(y) / log(x);

    // Compare with rounded value using a small 
    // tolerance to avoid floating point errors
    return fabs(res - round(res)) < 1e-10;
}

int main()
{
    printf("%s\n", isPower(10, 1) ? "true" : "false");
    printf("%s\n", isPower(1, 20) ? "true" : "false");
    printf("%s\n", isPower(2, 128) ? "true" : "false");
    printf("%s\n", isPower(2, 30) ? "true" : "false");

    return 0;
}
Java 
import java.lang.Math;

class GfG {

    static boolean isPower(int x, int y)
    {
        // Edge case: 1^k = 1 only
        if (x == 1)
            return y == 1;

        // Edge case: x^0 = 1
        if (y == 1)
            return true;

        // Compute logarithm
        double res = Math.log(y) / Math.log(x);

        // Compare with rounded value using a small
        // tolerance to avoid floating point errors
        return Math.abs(res - Math.round(res)) < 1e-10;
    }

    public static void main(String[] args)
    {
        System.out.println(isPower(10, 1) ? "true" : "false");
        System.out.println(isPower(1, 20) ? "true" : "false");
        System.out.println(isPower(2, 128) ? "true" : "false");
        System.out.println(isPower(2, 30) ? "true" : "false");
    }
}
Python 
import math

def isPower(x, y):

    # Edge case: 1^k = 1 only
    if x == 1:
        return y == 1

    # Edge case: x^0 = 1
    if y == 1:
        return True

    # Compute logarithm
    res = math.log(y) / math.log(x)

    # Compare with rounded value using a small 
    # tolerance to avoid floating point errors
    return abs(res - round(res)) < 1e-10


if __name__ == "__main__":
    print("true" if isPower(10, 1) else "false")
    print("true" if isPower(1, 20) else "false")
    print("true" if isPower(2, 128) else "false")
    print("true" if isPower(2, 30) else "false")
C# 
using System;

class GfG {

    static bool isPower(int x, int y)
    {
        // Edge case: 1^k = 1 only
        if (x == 1)
            return y == 1;

        // Edge case: x^0 = 1
        if (y == 1)
            return true;

        // Compute logarithm
        double res = Math.Log(y) / Math.Log(x);

        // Compare with rounded value using a small
        // tolerance to avoid floating point errors
        return Math.Abs(res - Math.Round(res)) < 1e-10;
    }

    static void Main()
    {
        Console.WriteLine(isPower(10, 1) ? "true" : "false");
        Console.WriteLine(isPower(1, 20) ? "true" : "false");
        Console.WriteLine(isPower(2, 128) ? "true" : "false");
        Console.WriteLine(isPower(2, 30) ? "true" : "false");
    }
}
JavaScript 
function isPower(x, y)
{
    // Edge case: 1^k = 1 only
    if (x === 1)
        return y === 1;

    // Edge case: x^0 = 1
    if (y === 1)
        return true;

    // Compute logarithm
    let res = Math.log(y) / Math.log(x);

    // Compare with rounded value using a small tolerance to
    // avoid floating point errors
    return Math.abs(res - Math.round(res)) < 1e-10;
}

// Driver code
console.log(isPower(10, 1) ? "true" : "false");
console.log(isPower(1, 20) ? "true" : "false");
console.log(isPower(2, 128) ? "true" : "false");
console.log(isPower(2, 30) ? "true" : "false");

Output
true
false
true
false
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Article Tags:
Mathematical
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