19. \( \frac{1}{2} \cdot(\ln 64+5 \cdot \ln k)-4 \cdot \ln m \)

To rewrite the expression \( \frac{1}{2} \cdot(\ln 64+5 \cdot \ln k)-4 \cdot \ln m \), we first simplify it step by step. We know from logarithmic properties that \( \frac{1}{2} \cdot \ln A = \ln(A^{1/2}) \) and \( a \cdot \ln B = \ln(B^a) \). Start with the first part: \[ \frac{1}{2} \cdot (\ln 64 + 5 \cdot \ln k) = \frac{1}{2} \cdot \ln 64 + \frac{5}{2} \cdot \ln k = \ln(64^{1/2}) + \ln(k^{5/2}) = \ln(8) + \ln(k^{5/2}) = \ln(8k^{5/2}). \] Now substitute this back into the expression: \[ \ln(8k^{5/2}) - 4 \cdot \ln m. \] Using the property \( \ln A - \ln B = \ln \left( \frac{A}{B} \right) \), we have: \[ \ln \left( \frac{8k^{5/2}}{m^4} \right). \] So, the final simplified expression becomes: \[ \ln \left( \frac{8k^{5/2}}{m^4} \right). \]