Expand the logarithm as much as possible. Rewrite the expression as a sum, difference, or product of logs.

 

ln(1/49^k)

To expand the logarithmic expression \( \ln\left(\frac{1}{49^k}\right) \), we can use the properties of logarithms.

1. **Quotient Rule**: The logarithm of a quotient is the difference of the logarithms:
   \[
   \ln\left(\frac{1}{a}\right) = -\ln(a)
   \]
   So, applying this to our expression:
   \[
   \ln\left(\frac{1}{49^k}\right) = -\ln\left(49^k\right)
   \]

2. **Power Rule**: The logarithm of a power is the exponent times the logarithm of the base:
   \[
   \ln(a^b) = b \cdot \ln(a)
   \]
   Applying this rule:
   \[
   -\ln\left(49^k\right) = -k \cdot \ln(49)
   \]

Thus, the expanded form of \( \ln\left(\frac{1}{49^k}\right) \) is:
\[
-k \cdot \ln(49)
\]

Would you like further details on logarithmic properties or any other questions?

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**Related questions:**
1. What is the logarithmic form of \( \ln\left(\frac{1}{x^2}\right) \)?
2. How would you expand \( \ln\left(\frac{25}{49^k}\right) \)?
3. How do you simplify \( \ln(a^b \cdot c^d) \)?
4. What is the derivative of \( \ln\left(\frac{1}{49^k}\right) \)?
5. How do you solve \( \ln(49^k) = 3 \) for \( k \)?

**Tip:** Always remember the quotient, product, and power rules when working with logarithmic expansions!

Expand the logarithm as much as possible. Rewrite the expression as a sum, difference, or product of logs. ln(1/49^k)

Learn how to expand the logarithmic expression ln(1/49^k) using the quotient and power rules. This problem demonstrates key logarithmic properties and provides a detailed step-by-step solution. Ideal for high school students in Grades 9-12.

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