\(\log_3 ( x^2 ) = \log_3 ( 3x + 2 )\), so \(x^2 = 3x + 2\). Thus \(x^2 - 3x - 2 = 0\). This expression does not factor, so we use the quadratic formula to solve for \(x\): \(x=\dfrac{3\pm\sqrt{17}}{2}\)
We must be cautious, however, since we can only take logarithms of positive numbers, so both \(x^2\) and \(3x+2\) must be positive. The positive solution \(x=\dfrac{3+\sqrt{17}}{2}\) satisfies both requirements, but it is not clear whether the negative solution \(x=\dfrac{3-\sqrt{17}}{2}\) satisfies the requirement that \(3x+2\) be positive, so it must be checked. If \(x=\dfrac{3-\sqrt{17}}{2}\), then \(3x+2=\dfrac{13-3\sqrt{17}}{2}\) which is positive, since \(13\) is larger than \(3\sqrt{17}\). This can easily be checked by comparing their squares. The square of \(13\) is \(169\) and the square of \(3\sqrt{17}\) is \(153\). Thus both solutions are valid.