# 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:

1. Compute the log of N in base K (logK N) to get the exponential power such that K raised to this exponent is the Highest power of K less than equal to N.
2. 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++ implementation of the approach`   `#include ` `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;` `}`

 `// 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`

 `# 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`

 `// C# implementation of the approach` `using` `System;` `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)` `{` `    ``Console.WriteLine(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 gauravrajput1`

Output:
```9 27

```

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