To prove the logarithmic identity:

$\log_a(xy) = \log_a x + \log_a y$

Proof:

1. Start with the definition of logarithms: A logarithm is the inverse of exponentiation. By definition, if:

   $\log_a x = m \quad \text{then} \quad a^m = x$ and $\log_a y = n \quad \text{then} \quad a^n = y$

2. Multiplying the arguments of the logarithms: From the expressions above, multiplying $x$ and $y$ gives: $x \cdot y = a^m \cdot a^n$

3. Apply exponentiation property: Using the property of exponents $a^m \cdot a^n = a^{m+n}$, we get: $x \cdot y = a^{m+n}$

4. Take the logarithm of both sides: Taking the logarithm base $a$ of both sides, we get: $\log_a(x \cdot y) = \log_a(a^{m+n})$

5. Use the inverse property of logarithms: Since $\log_a(a^k) = k$, the right-hand side simplifies to: $\log_a(x \cdot y) = m + n$

6. Substitute back the values of $m$ and $n$: Recall that $m = \log_a x$ and $n = \log_a y$, so we get: $\log_a(x \cdot y) = \log_a x + \log_a y$

This completes the proof.

Would you like further details or have any other questions?

Here are 5 related questions for deeper understanding:

1. How does this property of logarithms help in simplifying complex logarithmic expressions?
2. Can you prove similar properties for logarithmic division?
3. How can you use this property when solving exponential equations?
4. What is the geometric interpretation of logarithms?
5. How does changing the base $a$ affect this property of logarithms?

Tip: The logarithmic multiplication property is very useful in many areas of mathematics, especially when simplifying multiplication of large numbers or solving equations involving exponents.