Solve the equation 2log(3) (x-3) + log(1/3) (x + 1) = 2?

Question

Note: (3) and (1/3) after log, are the base (dunno if thats what you call it, cant remember now)..like log is usually to base 10, so log(10), here it's log(3) and log (1/3).

plz explain steps more clearly, i am not following

Answer 1

2 log[3](x-3) - log[3](x+1) = 2
(x-3)^2/(x+1) = 9
x^2 - 6x+9 = 9x+9
x^2 = 15x
so x= 0 :: disallow because of log(x-3) above
or x = 15 is your answer

Answer 2

log(3)(x+1) =log(1/3)(x+1) /log(1/3) (3).
The denominator is -1 as (1/3)^-1=3
So working in base 3 we get
log(x-3)^2-log(x+1)=2
so log (x-3)^2/(x+1) = 2 and (x-3)^2/(x+1)=3^2
x^2-6x+9 =9x+9
x^2-15x=0 x(x-15)=0 x=0 can´t be because as negative numbers don´have log x must be >3
x=15 is the solution