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!