To solve the given problem, we need to combine the expression

$\log_8 2 + \log_8 3 + \log_8 (x^2 - 1)$

into a single logarithm.

Solution Steps

1. Apply the Product Rule for Logarithms: Recall that the sum of logarithms with the same base can be combined using the product rule: $\log_b a + \log_b c = \log_b (a \cdot c)$

2. Combine the Terms: Using the product rule, we can rewrite the expression as: $\log_8 2 + \log_8 3 + \log_8 (x^2 - 1) = \log_8 (2 \cdot 3 \cdot (x^2 - 1))$

3. Simplify the Product Inside the Logarithm: Multiply the terms inside the logarithm: $\log_8 (2 \cdot 3 \cdot (x^2 - 1)) = \log_8 (6(x^2 - 1))$

4. Final Answer: The expression simplifies to: $\log_8 (6(x^2 - 1))$

So, the answer is: $\log_8 (6(x^2 - 1))$

Would you like further clarification on this solution?

Here are five related questions to deepen your understanding:

1. How would you simplify $\log_5 4 + \log_5 7$ into a single logarithm?
2. What is the result of $\log_8 (x^2 - 1)$ if $x = 2$?
3. How would you approach a problem that includes subtraction of logs, such as $\log_8 10 - \log_8 2$?
4. What is the change of base formula for logarithms, and how would it apply to base 8?
5. Can this expression, $\log_8 (6(x^2 - 1))$, be further simplified if $x^2 - 1$ was a specific number?

Tip: When combining logarithmic expressions, always look for opportunities to apply the product, quotient, and power rules of logarithms to simplify expressions.