The image shows the problem:

Show that: $\log_{16}(xy) = \frac{1}{2}\log_{4}(x) + \frac{1}{2}\log_{4}(y)$

Solution:

We begin by expressing the left-hand side, $\log_{16}(xy)$, using properties of logarithms and simplifying it.

Step 1: Change of Base Formula

First, recall that: $\log_{a}(b) = \frac{\log(b)}{\log(a)}$ Using this for $\log_{16}(xy)$, we get: $\log_{16}(xy) = \frac{\log(xy)}{\log(16)}$ But $\log(16)$ can be simplified because $16 = 4^2$, so: $\log(16) = \log(4^2) = 2\log(4)$ Thus: $\log_{16}(xy) = \frac{\log(xy)}{2\log(4)}$

Step 2: Apply Logarithmic Properties

Now, use the logarithmic property $\log(ab) = \log(a) + \log(b)$ to simplify $\log(xy)$: $\log(xy) = \log(x) + \log(y)$ Substitute this into the equation: $\log_{16}(xy) = \frac{\log(x) + \log(y)}{2\log(4)}$

Step 3: Separate the Terms

This can now be written as: $\log_{16}(xy) = \frac{1}{2}\left(\frac{\log(x)}{\log(4)} + \frac{\log(y)}{\log(4)}\right)$ Notice that $\frac{\log(x)}{\log(4)} = \log_{4}(x)$ and $\frac{\log(y)}{\log(4)} = \log_{4}(y)$, so: $\log_{16}(xy) = \frac{1}{2}\log_{4}(x) + \frac{1}{2}\log_{4}(y)$

Thus, we have shown that: $\log_{16}(xy) = \frac{1}{2}\log_{4}(x) + \frac{1}{2}\log_{4}(y)$

Would you like further details or explanations on any step?

Here are 5 related questions to explore:

1. What is the change of base formula and how is it used in logarithmic equations?
2. How can logarithmic properties help simplify expressions involving products?
3. Can we generalize this result for different bases, such as $\log_b(a)$?
4. What happens if we change the base of logarithms from base 16 to base 2 in this problem?
5. How can logarithms be applied to solve exponential equations?

Tip: Always remember that logarithms convert multiplication into addition, making it easier to handle complex expressions!