OpenSAT

Using the property of logarithms that states $\log_a b + \log_a c = \log_a (b \cdot c)$, we can simplify the equation: $\log_3 [(x+2)(x-1)] = 2$. Converting to exponential form, we get $3^2 = (x+2)(x-1)$. Expanding the right side of the equation gives $9 = x^2 + x - 2$. Subtracting 9 from both sides yields $x^2 + x -11 = 0$. This quadratic equation can be solved using the quadratic formula. In the quadratic formula, we have $a = 1$, $b = 1$, and $c = -11$. Plugging these values into the quadratic formula gives: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -11}}{2 \cdot 1}$. Simplifying this expression gives $x = \frac{-1 \pm \sqrt{45}}{2}$, or $x = \frac{-1 \pm 3\sqrt{5}}{2}$. Only $x = \frac{-1 + 3\sqrt{5}}{2}$ is given as a choice, which is approximately 4.