To express $\log_6(36 \cdot 25)$ as a sum of logarithms, we can apply the logarithmic property for products:

$\log_b(M \cdot N) = \log_b(M) + \log_b(N)$

In this case: $\log_6(36 \cdot 25) = \log_6(36) + \log_6(25)$

So, the answer is:

$\log_6(36) + \log_6(25)$

Would you like a further breakdown of this solution, or have any other questions?

Here are related questions you might find helpful:

1. How would this approach change if it were a division inside the logarithm?
2. What if the base of the logarithm was changed from 6 to 10?
3. Can this be simplified further by expressing $36$ and $25$ as powers of smaller numbers?
4. How do we handle the sum of logarithms if they have different bases?
5. What are the applications of logarithmic properties like this in real-world problems?

Tip: Remember, the properties of logarithms (product, quotient, and power rules) are powerful tools for simplifying expressions and solving equations.