Expand the logarithm as much as possible. Rewrite the expression as a sum, difference, or product of logs. \[ \ln \left(\frac{1}{9^{k}}\right) \] Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example, \( c^{*} \ln (h) \). \( \ln \left(\frac{1}{9^{k}}\right)= \)

To expand the logarithm \(\ln \left(\frac{1}{9^{k}}\right)\) as much as possible, we can use the properties of logarithms step by step. 1. **Quotient Rule of Logarithms:** \[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \] Apply this to the given expression: \[ \ln \left(\frac{1}{9^{k}}\right) = \ln(1) - \ln(9^{k}) \] 2. **Simplify \(\ln(1)\):** \[ \ln(1) = 0 \] So the expression becomes: \[ 0 - \ln(9^{k}) = -\ln(9^{k}) \] 3. **Power Rule of Logarithms:** \[ \ln(a^{b}) = b \cdot \ln(a) \] Apply this to \(\ln(9^{k})\): \[ -\ln(9^{k}) = -k \cdot \ln(9) \] 4. **Expressing 9 as a Power of 3:** Since \(9 = 3^{2}\), we can further expand \(\ln(9)\): \[ \ln(9) = \ln(3^{2}) = 2 \cdot \ln(3) \] Substitute this back into the expression: \[ -k \cdot \ln(9) = -k \cdot 2 \cdot \ln(3) = -2k \cdot \ln(3) \] **Final Expanded Form:** \[ \ln \left(\frac{1}{9^{k}}\right) = -2 \cdot k \cdot \ln(3) \]