Let's go through the problems using logarithmic properties.

Problem (a):

You are asked to expand the logarithmic expression:

$\ln\left( \sqrt{\frac{8x}{x - 8}} \right)$

Step 1: Use the square root property.

$\ln\left( \sqrt{\frac{8x}{x - 8}} \right) = \frac{1}{2} \ln\left( \frac{8x}{x - 8} \right)$

Step 2: Apply the logarithm of a quotient rule.

$\frac{1}{2} \ln\left( \frac{8x}{x - 8} \right) = \frac{1}{2} \left( \ln(8x) - \ln(x - 8) \right)$

Step 3: Expand the product $\ln(8x)$ using the logarithmic product rule.

$\frac{1}{2} \left( \ln(8x) - \ln(x - 8) \right) = \frac{1}{2} \left( \ln(8) + \ln(x) - \ln(x - 8) \right)$

This is the correct expanded form for part (a).

Problem (b):

You are asked to expand:

$\log_2 \left[ (x^3 + 2)^{\frac{2}{3}} (x - 7)^3 \right]$

Step 1: Use the logarithmic property of a product.

$\log_2 \left[ (x^3 + 2)^{\frac{2}{3}} (x - 7)^3 \right] = \log_2 \left( (x^3 + 2)^{\frac{2}{3}} \right) + \log_2 \left( (x - 7)^3 \right)$

Step 2: Apply the logarithmic power rule to both terms.

$= \frac{2}{3} \log_2(x^3 + 2) + 3 \log_2(x - 7)$

This is the expanded form for part (b).

Corrections:

• (a): The student's mistake seems to be in the coefficient handling; the final form is correct now.
• (b): The student appears to have used incorrect logarithm bases (log_3 instead of log_2). The correct expanded form should use log base 2 as per the original expression.

Would you like further clarification on any step or additional details?

Related Questions:

1. How do you simplify logarithmic expressions involving fractional exponents?
2. What is the quotient rule for logarithms, and how is it applied?
3. Can you explain how the product rule for logarithms is derived?
4. What are common mistakes to avoid when expanding logarithms?
5. How do you convert logarithmic expressions between different bases?

Tip:

Always double-check the logarithmic base in each expression, as using the wrong base can lead to incorrect answers.