We are asked to express the given logarithmic expression in the form $\log_r a$, and simplify where possible.

The expression is:

$\frac{1}{4} \log_r 16 + \frac{1}{2} \log_r 9 - 2 \log_r 4$

Step 1: Use logarithmic power rule

We apply the power rule of logarithms: $\frac{1}{n} \log_r x = \log_r x^{1/n}$, or $\log_r x^n = n \log_r x$. Each coefficient in front of the logarithms can be moved inside as an exponent.

$\frac{1}{4} \log_r 16 = \log_r 16^{1/4} = \log_r 2$ (because $16^{1/4} = 2$)

$\frac{1}{2} \log_r 9 = \log_r 9^{1/2} = \log_r 3$ (because $9^{1/2} = 3$)

$2 \log_r 4 = \log_r 4^2 = \log_r 16$

Step 2: Combine the terms

Now, we combine the logarithmic terms using the addition and subtraction properties of logarithms: $\log_r a + \log_r b = \log_r (a \cdot b)$ and $\log_r a - \log_r b = \log_r \left(\frac{a}{b}\right)$.

So, combining the terms:

$\log_r 2 + \log_r 3 - \log_r 16 = \log_r \left(\frac{2 \cdot 3}{16}\right)$

Step 3: Simplify the fraction

$\frac{2 \cdot 3}{16} = \frac{6}{16} = \frac{3}{8}$

Thus, the expression becomes:

$\log_r \frac{3}{8}$

Final Answer:

The simplified form of the given expression is:

$\log_r \frac{3}{8}$

Let me know if you want more details or further explanation.

Here are five related questions:

1. How does the power rule of logarithms work?
2. Can you apply logarithmic properties when simplifying without converting coefficients to exponents?
3. What is the addition property of logarithms, and when do you use it?
4. How do you handle subtraction of logarithms involving different numbers?
5. How can you check if your simplified logarithmic expression is correct?

Tip: When simplifying logarithms, always look for opportunities to apply power, product, and quotient rules to combine terms efficiently!