We are given the following expression to simplify:

$\log_b \left(\frac{8}{27}\right) - 2 \log_b \left(\frac{4}{3}\right) + 2 \log_b \left(\frac{3}{2}\right)$

Let's break this down step by step:

Step 1: Use the logarithmic power rule

Recall that $n \log_b x = \log_b (x^n)$. We apply this to the second and third terms: $\log_b \left(\frac{8}{27}\right) - \log_b \left(\left(\frac{4}{3}\right)^2\right) + \log_b \left(\left(\frac{3}{2}\right)^2\right)$

Step 2: Simplify the exponents

Now simplify the terms inside the logarithms: $\log_b \left(\frac{8}{27}\right) - \log_b \left(\frac{16}{9}\right) + \log_b \left(\frac{9}{4}\right)$

Step 3: Combine logarithms using the logarithmic difference rule

Now, use the rule $\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right)$ to combine the first two terms: $\log_b \left(\frac{8}{27} \div \frac{16}{9}\right) + \log_b \left(\frac{9}{4}\right)$

Step 4: Simplify the fractions

First, simplify the division of fractions: $\frac{8}{27} \div \frac{16}{9} = \frac{8}{27} \times \frac{9}{16} = \frac{72}{432} = \frac{1}{6}$ So the expression becomes: $\log_b \left(\frac{1}{6}\right) + \log_b \left(\frac{9}{4}\right)$

Step 5: Use the logarithmic addition rule

Now apply $\log_b A + \log_b B = \log_b (A \times B)$: $\log_b \left(\frac{1}{6} \times \frac{9}{4}\right)$

Step 6: Simplify the product

Now, simplify the product inside the logarithm: $\frac{1}{6} \times \frac{9}{4} = \frac{9}{24} = \frac{3}{8}$

Final Answer:

The simplified expression is: $\log_b \left(\frac{3}{8}\right)$

Would you like further details or have any questions about the steps?

Here are five related questions:

1. How does the logarithmic power rule apply to logarithmic expressions?
2. What is the difference between logarithmic addition and subtraction rules?
3. How can simplifying fractions help in logarithmic expressions?
4. How do you apply logarithmic rules in solving equations?
5. Can you explain how to interpret logarithms with different bases?

Tip: Always check if the logarithmic base is explicitly given. If not, assume it is a general base $b$, as done here.