Patrick B. answered 08/11/20
Math and computer tutor/teacher
Set up an equation B^x = n where B is the base and n is the number.
If the solution is irrational then so is the number and the decimal will recur.
Ex. 1/6 as base 8
8^x = 1/6
(2^3)^x = 6^(-1)
2^(3x) = 6^(-1)
takes log of both sides:
(3x) log 2 = - log 6
3x = - log 6/log2
x = -log 6 / (3 log 2)
severely irrational, decimal recurs