Question

Write a function named bytwo whose input is a positive number x. It will return the number of times x must be multiplied by 2 until the value becomes 100 or greater. If the input x is a negative number or 0, the function should return 0. Use a loop to count the number of times the value is multiplied by 2.

Answer 1

To solve this problem, we can use a loop to count the number of times we can multiply the given positive number by 2 until it becomes 100 or greater.

Step-by-step explanation:

We can use a loop to count how many times we can multiply the given positive number by two until it is 100 or more in order to solve this problem. A variable count will be initialized to 0. It will then be multiplied by two until it reaches 100 or higher. We will add one to the count each time we multiply. We will return 0 if the input number is negative or 0. The function's implementation is shown below:

def bytwo(x):
if x <= 0:
return 0
count = 0
while x < 100:
x *= 2
count += 1
return count