Question 1026110
log3 (x+1) -log3 (3x^2-3x-6)+log3 (x-2)


<i>The 3 on each log is lowered</i>
The 3 next to each "log" is the base for the logarithm.


log(3,(x+1))-log(3,(3x^2-3x-6))+log(3,(x-2))
{{{log(3,(x+1))-log(3,(3x^2-3x-6))+log(3,(x-2))}}}


{{{log(3,((x+1)(x-2)/(3x^2-3x-6)))}}}