Its very simple indeed... I bet you stopped trying when you saw the x³....if you continue with the simplifications you end up with a easy quadratic.
Since we have all logs @ same base we don't need the info "using a base of 3"
And because I'm a bit lazy today (:|), let l(u) denote the logarithm of the expression u
2l(x+2) + l(x-1) = 3 l(x)
//I assume you know how to manipulate logs.
l[ (x+2)² ] + l(x-1) = l(x³)
l[ (x+2)².(x-1) ] = l(x³)
//if we have l(z)=l(w) we can solve [z=w] with [z and w >0]
(x+2)².(x-1) = x³; x>1
//now this is the fun part
(x+2)².(x-1) = x³
(x²+4x+4).(x-1) = x³
x³+4x²+4x-x²-4x-4 = x³
x³+4x²-x²-4 = x³
+x³+3x²-4 = +x³
3x²-4 = 0
//I told you it was easy!
3x²-4 = 0
3x² = 4
x² = 4/3
x = ±(2√3)/3
//remember the original equation's condition?? X>1
//we discard the negative solution, so the answer to the problem is
//omg...solving problems like these one makes me remember how I deeply miss my good ol' math classes.;|