Sum and product with log

Pretty sure there are much simpler ways to do this, but here's my first (successful) approach:

For the working below, treat $x$ as $u$ (I don't know why I used $x$ )

$\begin{aligned} 4(x+y) - 5xy & = 4(\log_{18} 36+\log_{24} 72) - 5 \log_{18} 36\log_{24} 72 \\ & = 4 \left(\frac {\log_6 36}{\log_6 18} + \frac {\log_6 72}{\log_6 24}\right) - 5 \cdot \frac {\log_6 36}{\log_6 18} \cdot \frac {\log_6 72}{\log_6 24} \\ & = 4 \left(\frac 2{\log_6 3+1} + \frac {\log_6 2 + 2}{2\log_6 2 + 1} \right) - 5 \cdot \frac 2{\log_6 3+1} \cdot \frac {\log_6 2 + 2}{2\log_6 2 + 1} \\ & = \frac {4\log_6 2 \log_6 3+10 \log_6 2+8\log_6 3-4}{(2\log_6 2 + 1)(\log_6 3+1)} \\ & = \frac {4\log_6 2 \log_6 3+2 \log_6 2 + 8(\log_6 2+\log_6 3) -4}{2\log_6 2 \log_6 3 + 2\log_6 2 + \log_6 3+1} \\ & = \frac {4\log_6 2 \log_6 3+2 \log_6 2 + 4}{2\log_6 2 \log_6 3 + \log_6 2 +2} \\ & = \boxed 2 \end{aligned}$