The problem asks to:

1. Show that: $\log_c xy = \frac{1}{2} \log_4 x + \frac{1}{2} \log_4 y$
2. Solve the equation: $\log_{16} xy = \sqrt{3}$

Step 1: Proving the logarithmic equality

We begin with the left-hand side of the equation: $\log_c xy$ Using the property of logarithms: $\log_b (mn) = \log_b m + \log_b n$ This gives: $\log_c xy = \log_c x + \log_c y$

Now, we need to express $\log_c x$ and $\log_c y$ in terms of base 4. We use the change of base formula: $\log_b a = \frac{\log_k a}{\log_k b}$ So, using base 4: $\log_c x = \frac{\log_4 x}{\log_4 c}, \quad \log_c y = \frac{\log_4 y}{\log_4 c}$ Given that $c = 16$, and noting that $16 = 4^2$, we have: $\log_4 c = \log_4 16 = 2$

Thus, we can substitute this back: $\log_c x = \frac{1}{2} \log_4 x, \quad \log_c y = \frac{1}{2} \log_4 y$

Therefore, the expression becomes: $\log_c xy = \frac{1}{2} \log_4 x + \frac{1}{2} \log_4 y$ which is the desired result.

Step 2: Solving $\log_{16} xy = \sqrt{3}$

We are given: $\log_{16} xy = \sqrt{3}$ Using the change of base formula again, we express the logarithm with base 4: $\log_{16} xy = \frac{\log_4 xy}{\log_4 16}$ Since $\log_4 16 = 2$, we have: $\log_{16} xy = \frac{1}{2} \log_4 xy$ Thus, the equation becomes: $\frac{1}{2} \log_4 xy = \sqrt{3}$ Multiplying both sides by 2: $\log_4 xy = 2\sqrt{3}$

Now, we rewrite this as an exponential equation: $xy = 4^{2\sqrt{3}}$

Thus, $xy = 4^{2\sqrt{3}}$. This is the solution for $xy$.

Would you like further explanation or help with any specific part of this solution? Here are five related questions for practice:

1. How would you prove other logarithmic identities using the change of base formula?
2. What are the properties of logarithms and how are they used in solving logarithmic equations?
3. How would the solution change if the base were different, say $\log_8$ instead of $\log_{16}$?
4. How can you apply logarithmic equations to real-world exponential growth problems?
5. What is the geometric interpretation of logarithmic and exponential functions?

Tip: The change of base formula is particularly useful when solving logarithmic equations that involve non-standard bases. Practice applying it to simplify logarithms!