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?

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.