Logarithms are back!

We can re-write the given expression as

$\begin{array}{rl} (a+b)^2 +a(10-a) - b(10+b) &= \ \ 2ab + 10a - 10b \\ &= \ \ 2(a - 5)(b + 5) + 50 \end{array}$

We now rewrite the two logs with a common base. Since both given bases and both given arguments are of the form $2^m3^n$ , we choose base $2$ (base $3$ would work equally well.)

$\begin{array}{lcccccr} a = \log_{12} 18 = \dfrac{\log_2 18}{\log_2 12} = \dfrac{1 + 2 \log_2 3}{2 + \log_2 3} & & & \implies & & & a - 5 = \dfrac{\text{-}9 - 3 \log_2 3}{2 + \log_2 3} = \dfrac{\text{-}3 (3 + \log_2 3)}{2 + \log_2 3} \\ \\ b = \log_{24} 54 = \dfrac{\log_2 54}{\log_2 24} = \dfrac{1 + 3 \log_2 3}{3 + \log_2 3} & & & \implies & & & b + 5 = \dfrac{16 + 8 \log_2 3}{3 + \log_2 3} = \dfrac{8 (2 + \log_2 3)}{3 + \log_2 3} \end{array}$

Then, returning to the expression above,

$\begin{array}{rl} 2(a - 5)(b + 5) + 50 &= 2 \left[ \dfrac{\text{-}3 (3 + \log_2 3)}{2 + \log_2 3} \right] \left[ \dfrac{8 (2 + \log_2 3)}{3 + \log_2 3} \right] + 50 \\ \\ &= 2 (\text{-}3) (8) + 50 = \boxed{2} \end{array}$