problem maths?

Question

is it possible to solve the below question with log

3^x+1 + 3^x +3 ^x-1 is divisble by 13 for all integers n.

if so, plz solve it

Answer 1

(3^(x + 1) + 3^x + 3^(x - 1))/13
((3^x * 3) + 3^x + ((3^x)/3))/13
((9(3^x) + 3(3^x) + (3^x))/3) / 13
((3^x)(9 + 3 + 1))/3) / 13
(13(3^x))/3) / 13
(13(3^x))/3 / (13/1))
(13(3^x)/3) * (1/13)
(13(3^x))/(3 * 13)
(3^x)/3
(3^x)/(3^1)
3^(x - 1)

so for all values of 3^(x - 1), the problem is divisible by 13

Answer 2

i do not know if you have to solve it using log but here is a simple way of solution
3^(x+1) + 3^x +3 ^(x-1)
=3^(x-1)*9 + 3^(x-1)*3 + 3^(x-1)
=(9+3+1) * 3^(x-1)
=13 * 3^(x-1)

note that x has to be a positive integer

Answer 3

lets try

let 3^x = y
so x log3 = logy

so we have s = 3y + y + y/3 = y (13/3)

but y is divisible by 3 for x>0 and hence s is divisible by 13

hmmm.... there is no way we could use the log....!