$X = \log_{12} 18$ 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 $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$

Let $I=\dfrac{\log 18}{\log 12}\cdot\dfrac{\log 54}{\log 24}+5 \left( \dfrac{\log 18}{\log 12}-\dfrac{\log 54}{\log 24} \right)$. Also, let $\log 3=x $ and $\log 2=y$.
Then,
$$I= \frac{\log 3^2\cdot2}{\log 2^2\cdot 3}.\frac{\log 3^3 \cdot 2}{\log 2^3\cdot 3}+5\left(\frac{\log 3^2\cdot2}{\log 2^2\cdot3}-\frac{\log 3^3\cdot2}{\log 2^3\cdot3}\right)= \frac{2x+y}{2y+x}\cdot\frac{3x+y}{3y+x}+5 \left( \frac{2x+y}{2y+x}-\frac{3x+y}{3y+x} \right)$$
$$=\frac{6x^2+5xy+y^2+10x^2+35xy+15y^2-15x^2-35xy-10y^2}{(2y+x)(3y+x)}=\frac{x^2+5xy+6y^2}{x^2+5xy+6y^2}=1$$ 
