Thread: What kind is an algorithm that opens large natural number exponents?

Hello everybody!
This is my 1st question in this forum.
I have passed just high-school level mathematics, so I would appreciate an answer that is understandable to such educational level of a person.
Time to time they say in popular science math books, that a large exponent number is such and such in length, for example 3^7146 contains 1846678925 integers or so (this is not a real example, I just put those numbers in screen here). They may also tell us that the last integer is, say, "9".
I would like to know, what kind is an algorithm they really use in their computers? It is hard for me to believe that they use naive repetitive multiplications time after time?
My question consearns expecially integer or natural number base numbers and exponents.

You can determine the number of integers in 3^7146 by doing the following calculation:

3^7146 = 10^y

The value of 'y' rounded up to the next highest integer is the number of integers you are looking for.

3^7147 = [3^(log_3 10)] ^ y = 3^[log_3 10)*y]

So y = 7147/(log_3 10) = 7147/2.059 = 3409.986. Hence 3^7147 has 3410 digits.

As for what the last digit is, if you write out the powers of 3 you will see a pattern where the last digit cycles through 3, 9, 7, 1, ... Since 7147 is divisible by 4 with a remainder of 3, the last digit is 7.

Thank You for quick and helpful answer.
(My calculator says that log_3 10 is about 2,095..., not 2,059..., but I guess that is just a keyboard error?)

Originally Posted by kpkkpk

Thank You for quick and helpful answer.
(My calculator says that log_3 10 is about 2,095..., not 2,059..., but I guess that is just a keyboard error?)

Yes, keyboard error: log_3 10 = 2.095903....