To expand \( \ln\left(\frac{1}{25^k}\right) \), follow these steps:

1. Use the property of logarithms for division:
   \[
   \ln\left(\frac{1}{25^k}\right) = \ln(1) - \ln(25^k)
   \]
   
2. Simplify using the fact that \( \ln(1) = 0 \):
   \[
   \ln\left(\frac{1}{25^k}\right) = -\ln(25^k)
   \]
   
3. Apply the logarithmic power rule \( \ln(a^b) = b\ln(a) \):
   \[
   -\ln(25^k) = -k \ln(25)
   \]
   
So the expansion of \( \ln\left(\frac{1}{25^k}\right) \) is:
\[
\boxed{-k \ln(25)}
\]

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. Expand \( \ln\left(\frac{1}{x^2}\right) \).
2. Simplify \( \ln(10^n) \).
3. What is the value of \( \ln(e^x) \)?
4. How would you expand \( \ln\left(\frac{a}{b^2}\right) \)?
5. Solve \( \ln(100) \) using logarithmic properties.

**Tip:** Remember that \( \ln(1) = 0 \) for any logarithmic base, which simplifies many expressions!

Learn how to expand the logarithmic expression ln(1/25^k) using the logarithmic quotient rule and power rule. This step-by-step solution simplifies the expression to -k ln(25). Perfect for students in Grades 10-12 covering logarithmic properties and exponentiation.

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