Given two positive integers a and b, task is to find the number of digits in a^b (a raised to the power b).
Input: a = 2 b = 5 Output: no. of digits = 2 Explanation: 2^5 = 32 Hence, no. of digits = 2 Input: a = 2 b = 100 Output: no. of digits = 31 Explanation: 2^100 = 1.2676506e+30 Hence, no. of digits = 31
The number of digits in a^b can be calculated using the formula:
Number of Digits = 1 + b * (log10a)
When a number is divided by 10, it is reduced by 1 digit.
554 / 10 = 55, 55 / 10 = 5
Notice, 554 initially has 3 digits but after division there are 2 digits 55 and after further division there is only 1 digit 5. So it can be concluded that to count number of digits, how many times a number is divided by 10 to reach 1 needs to be calculated.
log base 10 of a number is the number of times a number needs to be divided by 10 to reach 1 but as 1 itself is not included in log base 10, 1 is added to get the number of digits.
Note: Floor value of b * (log10a) is taken.
Below is the implementation to calculate the number of digits in a^b.
no.of digits = 31
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