To solve the given equation:

$\log_3[(3x + 3)(x + 3)] = 2$

we need to follow these steps:

1. Rewrite the Equation Using Exponentiation: Since $\log_3(A) = B$ implies $A = 3^B$, we can rewrite the equation as: $(3x + 3)(x + 3) = 3^2$ Simplifying, we get: $(3x + 3)(x + 3) = 9$

2. Expand the Left Side: Now, expand $(3x + 3)(x + 3)$: $(3x + 3)(x + 3) = 3x^2 + 9x + 3x + 9 = 3x^2 + 12x + 9$ So the equation becomes: $3x^2 + 12x + 9 = 9$

3. Simplify the Equation: Subtract 9 from both sides: $3x^2 + 12x = 0$ Factor out $3x$: $3x(x + 4) = 0$

4. Solve for $x$: Set each factor to zero: $3x = 0 \Rightarrow x = 0$ $x + 4 = 0 \Rightarrow x = -4$

5. Verify the Solutions: Substitute $x = 0$ and $x = -4$ back into the original expression $(3x + 3)(x + 3)$ to ensure they do not produce negative values inside the logarithm (which would be invalid):

   • For $x = 0$: $(3(0) + 3)(0 + 3) = 3 \cdot 3 = 9$, which is valid.
   • For $x = -4$: $(3(-4) + 3)(-4 + 3) = (-12 + 3)(-1) = -9$, which is not valid for a logarithm.

   Therefore, $x = -4$ is not a valid solution.

6. Conclusion: The solution set is: $\{0\}$

Would you like further clarification or details on any of the steps?

Here are some related questions for further exploration:

1. What are the conditions for the existence of solutions in logarithmic equations?
2. How does the change of base formula work for logarithmic functions?
3. Can you explain the process of expanding expressions in polynomial equations?
4. What other types of equations can be solved by converting logarithmic forms?
5. Why is it important to check for valid solutions in logarithmic equations?

Tip: Always check for extraneous solutions, especially in logarithmic equations, since the argument of a logarithm must be positive.