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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!

Solve for all values of x: log_6(2x^2 - 3) - log_6(3x + 2) = 0

Learn how to solve the logarithmic equation log_6(2x^2 - 3) - log_6(3x + 2) = 0 using logarithmic properties and the quadratic formula. This detailed step-by-step solution covers key algebraic and logarithmic concepts. Suitable for students in Grades 9-12.

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