To expand the expression $\ln \frac{x^2 - 1}{x^7}$ using the properties of logarithms, follow these steps:

1. Use the Quotient Rule of Logarithms: The rule states that $\ln \frac{A}{B} = \ln A - \ln B.$ Applying this to our expression: $\ln \frac{x^2 - 1}{x^7} = \ln(x^2 - 1) - \ln(x^7).$

2. Simplify $\ln(x^7)$ using the Power Rule of Logarithms: The Power Rule states that $\ln(x^k) = k \cdot \ln(x).$ Applying this to $\ln(x^7)$: $\ln(x^7) = 7 \ln(x).$ Substituting back, we get: $\ln(x^2 - 1) - \ln(x^7) = \ln(x^2 - 1) - 7 \ln(x).$

3. Further Expand $\ln(x^2 - 1)$ using the Difference of Squares: Notice that $x^2 - 1 = (x + 1)(x - 1).$ Using the Product Rule of Logarithms, which states that $\ln(AB) = \ln(A) + \ln(B)$, we can expand $\ln(x^2 - 1)$ as follows: $\ln(x^2 - 1) = \ln((x + 1)(x - 1)) = \ln(x + 1) + \ln(x - 1).$

4. Combine Everything: Substitute the expanded form of $\ln(x^2 - 1)$ into the expression: $\ln \frac{x^2 - 1}{x^7} = \ln(x + 1) + \ln(x - 1) - 7 \ln(x).$

Final Answer:

$\ln \frac{x^2 - 1}{x^7} = \ln(x + 1) + \ln(x - 1) - 7 \ln(x).$

Would you like a deeper explanation of any step?

Here are some related questions:

1. How does the Power Rule apply to logarithmic expressions?
2. What are the conditions under which the properties of logarithms apply?
3. Can this expression be simplified further if additional constraints are known?
4. How would the answer differ if the base of the logarithm changed from $\ln$ to $\log_{10}$?
5. How can logarithmic properties help in solving exponential equations?

Tip: Always check if an expression can be factored before applying logarithmic properties. Factoring can often make expansions simpler.