The given equation is:
$\log_3(3x + 2) - 2 = \log_3(-2x + 15)$

Solution:

1. Rewrite the equation:
   Start by isolating the logarithmic expressions: $\log_3(3x + 2) = \log_3(-2x + 15) + 2$

2. Simplify the constant term:
   Recall that $a \log_b(c) + k = \log_b(c) + \log_b(b^k)$. So the constant $+2$ can be written as $\log_3(3^2) = \log_3(9)$: $\log_3(3x + 2) = \log_3(-2x + 15) + \log_3(9)$

3. Use the logarithmic property $\log_b(A) + \log_b(B) = \log_b(A \cdot B)$:
   Combine the right-hand side: $\log_3(3x + 2) = \log_3(9(-2x + 15))$

4. Eliminate the logarithms:
   Since $\log_3(a) = \log_3(b)$ implies $a = b$, we equate the arguments: $3x + 2 = 9(-2x + 15)$

5. Expand and simplify the equation:
   Expand the right-hand side: $3x + 2 = -18x + 135$ Combine like terms: $3x + 18x = 135 - 2$ $21x = 133$

6. Solve for $x$: $x = \frac{133}{21} = \frac{19}{3}$

7. Check the solution:
   Substitute $x = \frac{19}{3}$ back into the original logarithmic expressions to ensure the arguments are positive:

   • $3x + 2 = 3\left(\frac{19}{3}\right) + 2 = 19 + 2 = 21 > 0$
   • $-2x + 15 = -2\left(\frac{19}{3}\right) + 15 = -\frac{38}{3} + \frac{45}{3} = \frac{7}{3} > 0$

   Since the arguments are positive, the solution is valid.

Final Answer:

$x = \frac{19}{3}$

Do you have any questions or want further clarification? Here are 5 related questions you might find helpful:

1. How do logarithmic properties simplify complex equations?
2. Why is it important to check the argument of a logarithm for positivity?
3. What are the key steps in solving logarithmic equations with multiple terms?
4. How can we handle logarithms with different bases?
5. What happens if the solution makes the logarithmic arguments negative?

Tip: Always verify solutions in logarithmic equations to ensure arguments remain within the valid domain ($> 0$).