hi!

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

x=(2√3)/3

//omg...solving problems like these one makes me remember how I deeply miss my good ol' math classes.;|