Given that $\log_x p = 9$ and $\log_x q = 6$, find the value of $\log_{\frac{p}{q}} x^2$. Solve the equation $\lg(3x + 1) + \lg(2x - 1) = 2 - \lg 2$ and explain why there is only one solution.

1. Apply the change of base formula to convert the logarithm to base $x$.

$\log_{\frac{p}{q}} x^2 = \frac{\log_x x^2}{\log_x \left(\frac{p}{q}\right)}$

2. Simplify the numerator and apply the quotient law of logarithms to the denominator.

$\frac{\log_x x^2}{\log_x \left(\frac{p}{q}\right)} = \frac{2}{\log_x p - \log_x q}$

3. Substitute the given values and evaluate.

$\frac{2}{9 - 6} = \frac{2}{3}$

1. Combine the logarithmic terms on the left-hand side.

$\lg(3x + 1) + \lg(2x - 1) = \lg((3x + 1)(2x - 1))$

2. Express the right-hand side as a single logarithm.

$2 - \lg 2 = \lg 100 - \lg 2 = \lg\left(\frac{100}{2}\right) = \lg 50$

3. Equate the arguments of the logarithms and expand.

$\lg((3x + 1)(2x - 1)) = \lg 50$
$(3x + 1)(2x - 1) = 50$
$6x^2 - 3x + 2x - 1 = 50$
$6x^2 - x - 51 = 0$

4. Solve the resulting quadratic equation.

$(6x + 17)(x - 3) = 0$
$x = -\frac{17}{6} \text{ or } x = 3$

5. Check the validity of the solutions and explain why there is only one solution.

When $x = -\frac{17}{6}$, $3x + 1 = 3\left(-\frac{17}{6}\right) + 1 = -7.5 < 0$.
Since the logarithm of a negative number is undefined, $x = -\frac{17}{6}$ is rejected.
Therefore, there is only one solution, $x = 3$.