How to check if an integer is a power of 3?

Question

I saw this question, and pop up this idea.

Answer 1

I find myself slightly thinking that if by 'integer' you mean 'signed 32-bit integer', then (pseudocode)

return (n == 1) 
    or (n == 3)
    or (n == 9)
    ... 
    or (n == 1162261467) 

has a certain beautiful simplicity to it (the last number is 3^19, so there aren't an absurd number of cases). Even for an unsigned 64-bit integer there still be only 41 cases (thanks @Alexandru for pointing out my brain-slip). And of course would be impossible for arbitrary-precision arithmetic...