1. Express \( a \) and \( b \) in Terms of Logarithms with Common Bases: - For \( a = \log_{12} 18 \): \[ a = \frac{\log 18}{\log 12} \] - For \( b = \log_{24} 54 \): \[ b = \frac{\log 54}{\log 24} \] 2. Simplify the Logarithms: - Express 18 and 54 in terms of their prime factors: \[ 18 = 2 \times 3^2 \quad \text{and} \quad 54 = 2 \times 3^3 \] - Express 12 and 24 in terms of their prime factors: \[ 12 = 2^2 \times 3 \quad \text{and} \quad 24 = 2^3 \times 3 \] 3. Calculate \( ab + 5(a - b) \): - Compute \( ab \): \[ ab = \left(\frac{\log 18}{\log 12}\right) \left(\frac{\log 54}{\log 24}\right) \] - Compute \( a - b \): \[ a - b = \frac{\log 18}{\log 12} - \frac{\log 54}{\log 24} \] - Combine the terms: \[ ab + 5(a - b) = \left(\frac{\log 18}{\log 12}\right) \left(\frac{\log 54}{\log 24}\right) + 5\left(\frac{\log 18}{\log 12} - \frac{\log 54}{\log 24}\right) \] 4. Simplify the Expression: - Notice that \( \log 18 = \log (2 \times 3^2) = \log 2 + 2 \log 3 \) - Similarly, \( \log 54 = \log (2 \times 3^3) = \log 2 + 3 \log 3 \) - Also, \( \log 12 = \log (2^2 \times 3) = 2 \log 2 + \log 3 \) - And \( \log 24 = \log (2^3 \times 3) = 3 \log 2 + \log 3 \) Substituting these into the expression and simplifying, we find that: \[ ab + 5(a - b) = 1 \] Therefore, the value of \( ab + 5(a - b) \) is: \[ \boxed{1} \]