how do i solve this equation?
log3(x^2)-log3(3x+2)=0
if i cube the logs does that undo them?
Rewrite the equation:
$\displaystyle \log_3(x^2)-\log_3(3x+2)=0~\implies~\log_3\left(\dfrac{x^2}{3x +2} \right)=0\ ,\ x > -\frac23$
Now use the LHS and the RHS as exponents to the base 3:
$\displaystyle \dfrac{x^2}{3x+2} = 1~\implies~x^2-3x-2=0$
This is a quadratic equation. Use the quadratic formula to solve this equation. You'll get 2 solutions, but be careful: One isn't valid. Why?