The expression \( \ln\left(\frac{1}{49^k}\right) \) can be simplified using logarithmic properties.

1. **Logarithm of a fraction**:
   \[
   \ln\left(\frac{1}{49^k}\right) = \ln(1) - \ln(49^k)
   \]
   Since \( \ln(1) = 0 \), this simplifies to:
   \[
   = -\ln(49^k)
   \]

2. **Logarithm of a power**:
   Now, apply the logarithmic rule \( \ln(a^b) = b \ln(a) \):
   \[
   -\ln(49^k) = -k \ln(49)
   \]

So, the simplified expression is:
\[
\ln\left(\frac{1}{49^k}\right) = -k \ln(49)
\]

Would you like further details or explanations on this?

Here are 5 related questions:
1. What are the basic properties of logarithms?
2. How do you apply the change of base formula for logarithms?
3. Can logarithms of negative numbers or zero be defined?
4. How do you differentiate and integrate logarithmic functions?
5. How do you solve logarithmic equations?

**Tip**: Always remember the key logarithmic identity \( \ln(a^b) = b \ln(a) \) when simplifying expressions with powers.

Learn how to simplify the logarithmic expression ln(1/49^k) by applying logarithmic properties. This problem demonstrates the use of the logarithm of a fraction and the logarithm of a power rule. Suitable for students in Grades 10-12.

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