We are tasked with simplifying the expression:

\[
2 \log 48 + 2 \log \sqrt{\frac{1}{16}} - 2 \log 3
\]

### Step 1: Apply the logarithmic rule \( a \log b = \log b^a \)

We start by applying the rule \( a \log b = \log b^a \) to each term that has a coefficient of 2:

\[
2 \log 48 = \log 48^2 = \log 2304
\]
\[
2 \log \sqrt{\frac{1}{16}} = \log \left( \sqrt{\frac{1}{16}}^2 \right) = \log \frac{1}{16}
\]
\[
2 \log 3 = \log 3^2 = \log 9
\]

So the expression now becomes:

\[
\log 2304 + \log \frac{1}{16} - \log 9
\]

### Step 2: Use the logarithmic property \( \log a + \log b = \log (a \cdot b) \)

Combine the first two logarithms:

\[
\log 2304 + \log \frac{1}{16} = \log \left( 2304 \times \frac{1}{16} \right) = \log 144
\]

Now the expression is:

\[
\log 144 - \log 9
\]

### Step 3: Apply the logarithmic property \( \log a - \log b = \log \left( \frac{a}{b} \right) \)

Finally, subtract the logs:

\[
\log \left( \frac{144}{9} \right) = \log 16
\]

### Final Answer:
\[
\log 16
\]

Let me know if you'd like further details or explanations!

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Here are 5 questions you might consider:
1. What is the logarithmic property used for addition and subtraction of logs?
2. How do you simplify logarithms involving square roots?
3. Can you apply the power rule of logarithms to more complex equations?
4. What is the significance of logarithmic bases in these types of problems?
5. How do you handle logarithmic expressions without numerical values?

**Tip:** Remember, whenever you see a coefficient in front of a logarithmic term, try applying the power rule \( a \log b = \log b^a \) for easier simplification!

Learn how to simplify the logarithmic expression ² log 48 + ² log √1/16 - ² log 3. This problem involves applying key logarithmic properties, including the power rule and product and quotient rules. Suitable for students in Grades 9-11.

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