The equation provided is:

$2 \log_6(x - 5) + \log_6(9) = 2$

Step-by-Step Solution:

1. Simplify using logarithmic properties: Recall the property $a \log_b(c) = \log_b(c^a)$. Apply this property to the first term: $2 \log_6(x - 5) = \log_6((x - 5)^2)$ So, the equation becomes: $\log_6((x - 5)^2) + \log_6(9) = 2$

2. Combine the logarithms: Use the property $\log_b(a) + \log_b(c) = \log_b(a \cdot c)$. Combine the two logarithmic terms: $\log_6((x - 5)^2 \cdot 9) = 2$

3. Rewrite the equation in exponential form: Recall that $\log_b(A) = C$ implies $A = b^C$. Rewrite the equation: $(x - 5)^2 \cdot 9 = 6^2$

4. Simplify: Compute $6^2 = 36$: $(x - 5)^2 \cdot 9 = 36$ Divide both sides by 9: $(x - 5)^2 = 4$

5. Solve for $x - 5$: Take the square root of both sides (consider both positive and negative roots): $x - 5 = \pm 2$

6. Solve for $x$: Add 5 to both sides: $x = 5 + 2 = 7 \quad \text{or} \quad x = 5 - 2 = 3$

7. Check for validity: The logarithmic argument $x - 5$ must be positive. For $x = 3$: $x - 5 = 3 - 5 = -2 \quad (\text{invalid, as logarithms of negative values are undefined}).$ Therefore, the only valid solution is: $x = 7$

Final Answer:

$x = 7$

Would you like additional clarification on any step? Here are five related questions to expand on:

1. How do the properties of logarithms simplify equations like this?
2. What happens if the base of the logarithm changes?
3. Why is $x = 3$ invalid in this context?
4. Can this type of equation be solved graphically?
5. How would the solution change if the base of the logarithm was unspecified?

Tip: Always check the domain of the logarithmic function to ensure the arguments are valid!