Write a function log2(x), which gives an integer approximation of the log base 2 of a positive number x

Write a function log2(x), which gives an integer approximation of the log base 2 of a positive number x, but it does so using recursion. Use a base case where you return 0 if x is 1 or less. In the general case, add 1 to the answer and divide x by 2. On top of writing the function, also write a test case for the function to prove the program works.

The Correct Answer and Explanation is :

To compute the integer approximation of the base-2 logarithm of a positive number x using recursion, we can define a function log2(x) that follows these steps:

1. Base Case: If x is less than or equal to 1, return 0. This is because the logarithm of 1 is 0, and for any number less than 1, the logarithm is negative, but since we’re approximating to the nearest integer, we return 0.
2. Recursive Case: If x is greater than 1, divide x by 2 and add 1 to the result of the recursive call. This effectively counts how many times x can be halved before it becomes less than or equal to 1, which corresponds to the integer part of the base-2 logarithm.

Here’s the Python implementation of the log2 function:

def log2(x):
    # Base case: if x is 1 or less, return 0
    if x <= 1:
        return 0
    # Recursive case: divide x by 2 and add 1
    else:
        return 1 + log2(x / 2)

Test Case:

To verify that the function works correctly, we can test it with a known value. For example, the base-2 logarithm of 8 is 3, since 8 = 2³. Let’s test the function with x = 8:

print(log2(8))  # Expected output: 3

Explanation:

• Base Case: When x is 1 or less, the function returns 0. This is because the logarithm of 1 is 0, and for any number less than 1, the logarithm is negative, but since we’re approximating to the nearest integer, we return 0.
• Recursive Case: For numbers greater than 1, the function recursively divides x by 2 and adds 1. This process effectively counts how many times x can be halved before it becomes less than or equal to 1, which corresponds to the integer part of the base-2 logarithm.

For example, with x = 8:

1. log2(8) calls log2(4) and adds 1.
2. log2(4) calls log2(2) and adds 1.
3. log2(2) calls log2(1) and adds 1.
4. log2(1) returns 0.

Adding up the results: 1 + 1 + 1 + 0 = 3, which is the correct integer approximation of the base-2 logarithm of 8.

This recursive approach effectively computes the integer part of the base-2 logarithm by counting how many times the number can be halved before it becomes less than or equal to 1.