$X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$ $X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way.
All I'm getting is $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$

$ X = \log_{12} 18 $ and  $ Y= \log_{24} 54 $.
$ X = \frac {\log_{2}18}{\log_{2}12}=\frac{\log_{2}9\cdot2}{\log_{2}4\cdot3}=\frac{\log_{2}3^2\cdot 2}{\log_{2}2^2\cdot3}=\frac{2\log_{2}3 + 1}{\log_{2}3 + 2}=\frac{2A+1}{A+2}$ 
$ Y = \frac {\log_{2}54}{\log_{2}18}=\frac{\log_{2}27\cdot2}{\log_{2}8\cdot3}=\frac{\log_{2}3^3\cdot 2}{\log_{2}2^3\cdot3}=\frac{2\log_{2}3 + 1}{\log_{2}3 + 2}=\frac{3A+1}{A+3}$, where $A=\log_{2}3$
By X have:
$X(A+2)=2A+1\Rightarrow AX+2X=2A+1 \Rightarrow A(X-2)=1-2X \Rightarrow A=\frac{1-2X}{X-2}$
courses, by Y have:
$Y(A+3)=A+1\Rightarrow Ay+3X=3A+1 \Rightarrow A(Y-3)=1-3XY\Rightarrow A=\frac{1-3Y}{Y-3}$
Here we have to:
$A=\frac{1-2X}{X-2}$, 
$A=\frac{1-3Y}{Y-3}$
where
$\frac{1-2X}{X-2}=\frac{1-3Y}{Y-3}$
$(1-2X)(Y-3)=(1-3Y)(X-2)$
$Y-2XY-3+6X=X-3XY-2+6Y$
$Y-2XY-3+6X-X+3XY+2-6Y=0$
$XY+5(X-Y)-1=0$
$XY+5(X-Y)=1$
