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Let's expand the expression step-by-step to solve it correctly.

The original expression is:

\[
\ln\left(\sqrt{7xy}\right)
\]

### Step 1: Use the Square Root Property
Rewrite the square root as a fractional exponent:
\[
\ln\left(\sqrt{7xy}\right) = \ln\left((7xy)^{\frac{1}{2}}\right)
\]

### Step 2: Apply the Power Rule for Logarithms
The power rule for logarithms states that \(\ln(a^b) = b \cdot \ln(a)\). So:
\[
\ln\left((7xy)^{\frac{1}{2}}\right) = \frac{1}{2} \cdot \ln(7xy)
\]

### Step 3: Expand \(\ln(7xy)\) Using the Product Rule
The product rule for logarithms states that \(\ln(abc) = \ln(a) + \ln(b) + \ln(c)\). Apply this to \(\ln(7xy)\):
\[
\ln(7xy) = \ln(7) + \ln(x) + \ln(y)
\]

### Step 4: Substitute and Simplify
Now, substitute back:
\[
\frac{1}{2} \cdot \left(\ln(7) + \ln(x) + \ln(y)\right)
\]

Distribute the \(\frac{1}{2}\):
\[
= \frac{1}{2} \ln(7) + \frac{1}{2} \ln(x) + \frac{1}{2} \ln(y)
\]

### Final Answer
This matches the correct choice:
\[
\frac{1}{2} \ln(7) + \frac{1}{2} \ln(x) + \frac{1}{2} \ln(y)
\]

This answer corresponds to the option:

- **Option 2: \( \frac{1}{2} [\ln(7) + \ln(x) + \ln(y)] \)**

Would you like more details on any of these steps?

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Here are some questions to further your understanding:
1. What is the difference between the product rule and the power rule for logarithms?
2. How would the steps change if it were \(\ln(\sqrt{7x/y})\) instead?
3. Why is it important to distribute the \(\frac{1}{2}\) across all terms in the logarithm?
4. How can the properties of logarithms help simplify complex expressions?
5. What is the effect of changing the base of the logarithm on this problem?

**Tip:** Always break down composite functions (like roots and products) into simpler parts using the rules of logarithms to make problems easier to handle.

Use rules of logarithms to expand \( \ln(\sqrt{7xy}) \).

Learn how to expand the logarithmic expression \( \ln(\sqrt{7xy}) \) using rules of logarithms. This solution covers the power and product rules, simplifying the expression step-by-step. Ideal for students in Grades 10-12.

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