We are given the expression $\ln\left( \frac{1}{25^k} \right)$ and asked to rewrite it using logarithmic properties. Let's break it down step by step:

Step 1: Apply the quotient rule for logarithms

The logarithmic quotient rule states: $\ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b)$ Using this rule, we can rewrite the expression: $\ln\left( \frac{1}{25^k} \right) = \ln(1) - \ln(25^k)$

Step 2: Simplify $\ln(1)$

Since $\ln(1) = 0$, we get: $0 - \ln(25^k) = -\ln(25^k)$

Step 3: Apply the power rule for logarithms

The power rule states: $\ln(a^b) = b \cdot \ln(a)$ Using this rule, we can rewrite $\ln(25^k)$ as: $\ln(25^k) = k \cdot \ln(25)$

Thus, the expression becomes: $-\ln(25^k) = -k \cdot \ln(25)$

Final Answer:

$\ln\left( \frac{1}{25^k} \right) = -k \cdot \ln(25)$

Would you like further details on any of the steps?

---

Here are five related questions to explore:

1. What is the value of $\ln(25)$ in terms of natural logarithms of prime factors?
2. How would the expansion change if the base of the logarithm were 10 instead of $\ln$?
3. How do you expand $\ln\left( \frac{x^2}{y^3} \right)$?
4. What is the derivative of $\ln(25^k)$ with respect to $k$?
5. Can you apply the same logarithmic rules to simplify $\ln\left( \frac{(x+2)}{y^k} \right)$?

Tip: Always remember the power rule in logarithms—it’s handy for simplifying powers inside a logarithmic expression!