**π‘ Problem Formulation:** We often encounter computational challenges where we need to verify if a certain number is a power of another. This article specifically tackles the problem of determining whether a given number is a power of d, given that d itself is a power of 2. For instance, with the input of number 64 and d = 4, the desired output would be `True`

, since 64 is 4^3, and 4 is 2^2.

## Method 1: Using Logarithms

This method applies the properties of logarithms to check if the given number is a power of another. It relies on the fact that if y is d to the power of x, the logarithm of y over the logarithm of d should be an integer. In Python, this can be efficiently computed using the `math`

library.

Here’s an example:

import math def is_power_of_d(num, d): return math.log(num, d).is_integer() print(is_power_of_d(64, 4))

Output: `True`

In this code snippet, we define a function `is_power_of_d`

that takes a number and a base d, calculates the logarithm of the number with base d, and checks if the result is an integer using the `is_integer()`

method.

## Method 2: Looping and Division

Looping and division is a straightforward method that keeps dividing the number by d until it cannot be divided anymore. If the final result is 1, the number is indeed a power of d; otherwise, it is not.

Here’s an example:

def is_power_of_d(num, d): while num % d == 0: num /= d return num == 1 print(is_power_of_d(64, 4))

Output: `True`

The `is_power_of_d`

function divides the given number by d continuously as long as the remainder is zero. The loop ends when the number cannot be evenly divided by d, at which point the function returns whether the remaining number is 1, indicating if the initial number was a power of d or not.

## Method 3: Bit Manipulation

This method utilizes bit manipulation to take advantage of the fact that powers of 2 have only one bit set in binary representation. We can, therefore, shift the bits to compare the powers directly.

Here’s an example:

def is_power_of_d(num, d): return num & (num - 1) == 0 and num % d == 0 print(is_power_of_d(64, 4))

Output: `True`

Here, `is_power_of_d`

first checks that the number is a power of 2 by ensuring that `num & (num - 1)`

equals 0. Then, it checks whether the number can be divided evenly by d.

## Method 4: Recursive Function

A recursive function successively reduces the problem size until it becomes trivial. This method applies the same division logic as Method 2, but through recursive function calls, checking the condition at each level of recursion.

Here’s an example:

def is_power_of_d(num, d): if num == 1: return True if num % d != 0: return False return is_power_of_d(num / d, d) print(is_power_of_d(64, 4))

Output: `True`

The function `is_power_of_d`

uses recursion to divide num by d, and at each step, it checks if the division results in a whole number. If num ever becomes 1, we know that it’s a power of d and return `True`

. If a non-integer division occurs, the function terminates and returns `False`

.

## Bonus One-Liner Method 5: Using Exponentiation

This concise one-liner uses exponentiation to check if raising d to an integer power can equal the number. It leverages list comprehension along with `any()`

function to assess the condition succinctly.

Here’s an example:

is_power_of_d = lambda num, d: any(num == d**i for i in range(int(math.log(num, 2)) + 1)) print(is_power_of_d(64, 4))

Output: `True`

The lambda function takes num and d, then iterates over a range of exponents up to the logarithm of num to the base 2. Using generator comprehension, it checks if `num == d**i`

for any i within the range, returning `True`

as soon as it finds a match.

## Summary/Discussion

**Method 1: Using Logarithms.**This method is mathematically elegant and has clear logic. However, it can be susceptible to floating-point precision errors for very large numbers.**Method 2: Looping and Division.**Straightforward and easy to understand. This method may be less efficient due to the looping, especially with very large numbers.**Method 3: Bit Manipulation.**Bitwise operations are usually very fast and this method highlights an optimized approach. It is, however, somewhat less readable and could be confusing for those unfamiliar with bit operations.**Method 4: Recursive Function.**Recursion can be an elegant solution, providing clear and concise code. The downside is the potential risk of hitting the recursion depth limit with exceptionally large inputs.**Bonus Method 5: Using Exponentiation.**The one-liner is quick and expressive. While it keeps the codebase small, the list comprehension might be computationally expensive for large numbers as it calculates powers until a match is found.