Pretty straightforward question. When I solved it, I got two positive and two negative solutions, so that would make 4 in total. None get discarded as the arguments in the logarithm still stay positive.
However, the solution is supposed to be 2, so I'm not sure where I went wrong.
I got the values of $x$ as $-2$, $1/2$, $12/5$ and $-18/5$
You need to use the change of base formula: $$ log_a{x}={\log_b{x}\over\log_b{a}}$$ Remove the polynomial from the base of the log on the right by changing the base to 3