The equation you've provided is:

$\log_6(2x^2 - 3) - \log_6(3x + 2) = 0$

Step-by-step solution:

1. Use the logarithmic property: The property $\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)$ allows us to combine the logarithms on the left-hand side: $\log_6\left(\frac{2x^2 - 3}{3x + 2}\right) = 0$

2. Rewrite the equation: By the definition of logarithms, $\log_b(a) = 0$ means that $a = 1$ (since $b^0 = 1$). Thus, we can set the argument of the logarithm equal to 1: $\frac{2x^2 - 3}{3x + 2} = 1$

3. Solve the rational equation: Multiply both sides by $3x + 2$ to eliminate the denominator: $2x^2 - 3 = 3x + 2$ Rearrange the equation: $2x^2 - 3x - 5 = 0$

4. Solve the quadratic equation: Use the quadratic formula to solve $2x^2 - 3x - 5 = 0$: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$ Simplify: $x = \frac{3 \pm \sqrt{9 + 40}}{4}$ $x = \frac{3 \pm \sqrt{49}}{4}$ $x = \frac{3 \pm 7}{4}$

   This gives two solutions: $x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5$ $x = \frac{3 - 7}{4} = \frac{-4}{4} = -1$

5. Check for extraneous solutions: Since logarithms are only defined for positive arguments, we must check that both solutions keep the arguments of the original logarithms positive:

   • For $x = 2.5$: $2(2.5)^2 - 3 = 12.5 - 3 = 9.5 > 0$ $3(2.5) + 2 = 7.5 + 2 = 9.5 > 0$ Both arguments are positive, so $x = 2.5$ is valid.

   • For $x = -1$: $2(-1)^2 - 3 = 2 - 3 = -1 < 0$ This makes the first logarithmic argument negative, which is invalid. So, $x = -1$ is not a valid solution.

Final Answer:

The only solution is $x = 2.5$.

Would you like further details or have any questions? Here are some related problems you might find interesting:

1. Solve $\log_3(x^2 - 4) - \log_3(x - 2) = 1$.
2. Solve $\log_2(5x + 6) - \log_2(x) = 3$.
3. Solve $\log_{10}(2x + 1) = 2$.
4. Solve $\log_7(x + 3) + \log_7(x - 1) = 1$.
5. Solve $\log_5(x^2 - 2x) - \log_5(x - 3) = 0$.

Tip: Always check the domain of logarithmic equations to ensure that the arguments are positive!