# Highest and Smallest power of K less than and greater than equal to N respectively

Given positive integers **N** and **K**, the task is to find the highest and smallest power of K greater than equal to and less than equal to N respectively.

**Examples:**

Input:N = 3, K = 2

Output:2 4

Highest power of 2 less than 3 = 2

Smallest power of 2 greater than 3 = 4

Input:N = 6, K = 3

Output:3 9

Highest power of 3 less than 6 = 2

Smallest power of 3 greater than 6 = 9

**Approach:**

- Compute the log of N in base K (
**log**) to get the exponential power such that K raised to this exponent is the Highest power of K less than equal to N._{K}N - For the Smallest power of K less than equal to N, find the next power of K computed from the last step

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the highest power ` `// of k less than or equal to n ` `int` `prevPowerofK(` `int` `n, ` `int` `k) ` `{ ` ` ` `int` `p = (` `int` `)(` `log` `(n) / ` `log` `(k)); ` ` ` `return` `(` `int` `)` `pow` `(k, p); ` `} ` ` ` `// Function to return the smallest power ` `// of k greater than or equal to n ` `int` `nextPowerOfK(` `int` `n, ` `int` `k) ` `{ ` ` ` `return` `prevPowerofK(n, k) * k; ` `} ` ` ` `// Function to print the result ` `void` `printResult(` `int` `n, ` `int` `k) ` `{ ` ` ` `cout << prevPowerofK(n, k) ` ` ` `<< ` `" "` `<< nextPowerOfK(n, k) ` ` ` `<< endl; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 25, k = 3; ` ` ` ` ` `printResult(n, k); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the approach ` `import` `java.io.*; ` ` ` `class` `GFG{ ` ` ` `// Function to return the highest power ` `// of k less than or equal to n ` `static` `int` `prevPowerofK(` `int` `n, ` `int` `k) ` `{ ` ` ` `int` `p = (` `int` `)(Math.log(n) / Math.log(k)); ` ` ` `return` `(` `int` `) Math.pow(k, p); ` `} ` ` ` `// Function to return the smallest power ` `// of k greater than or equal to n ` `static` `int` `nextPowerOfK(` `int` `n, ` `int` `k) ` `{ ` ` ` `return` `prevPowerofK(n, k) * k; ` `} ` ` ` `// Function to print the result ` `static` `void` `printResult(` `int` `n, ` `int` `k) ` `{ ` ` ` `System.out.println(prevPowerofK(n, k) + ` `" "` `+ ` ` ` `nextPowerOfK(n, k)); ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main (String args[]) ` `{ ` ` ` `int` `n = ` `25` `, k = ` `3` `; ` ` ` `printResult(n, k); ` `} ` `} ` ` ` `// This code is contributed by shivanisinghss2110 ` |

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## Python3

`# Python3 implementation of the approach ` `import` `math ` ` ` `# Function to return the highest power ` `# of k less than or equal to n ` `def` `prevPowerofK(n, k): ` ` ` ` ` `p ` `=` `int` `(math.log(n) ` `/` `math.log(k)) ` ` ` `return` `int` `(math.` `pow` `(k, p)) ` ` ` `# Function to return the smallest power ` `# of k greater than or equal to n ` `def` `nextPowerOfK(n, k): ` ` ` ` ` `return` `prevPowerofK(n, k) ` `*` `k ` ` ` `# Function to print the result ` `def` `printResult(n, k): ` ` ` ` ` `print` `(prevPowerofK(n, k), nextPowerOfK(n, k)) ` ` ` `# Driver code ` `n ` `=` `6` `k ` `=` `3` ` ` `printResult(n, k) ` ` ` `# This code is contributed by divyamohan123 ` |

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**Output:**

9 27

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