$X = \log_{12} 18$ 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 $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$
 A: 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,
$$I= \frac{\log 3^2\cdot2}{\log 2^2\cdot 3}.\frac{\log 3^3 \cdot 2}{\log 2^3\cdot 3}+5\left(\frac{\log 3^2\cdot2}{\log 2^2\cdot3}-\frac{\log 3^3\cdot2}{\log 2^3\cdot3}\right)= \frac{2x+y}{2y+x}\cdot\frac{3x+y}{3y+x}+5 \left( \frac{2x+y}{2y+x}-\frac{3x+y}{3y+x} \right)$$
$$=\frac{6x^2+5xy+y^2+10x^2+35xy+15y^2-15x^2-35xy-10y^2}{(2y+x)(3y+x)}=\frac{x^2+5xy+6y^2}{x^2+5xy+6y^2}=1$$ 
A: Note that $XY + 5(X - Y) = (X - 5)(Y + 5) + 25$, so it suffices to find $(X - 5)(Y + 5)$.
$(X - 5) = \log_{12}(18) - 5 = \log_{12}{18 \over 12^5} = \log_{12}{3^{-3}2^{-9}} = -3\log_{12}(24)$.
$(Y + 5) = \log_{24}(54) + 5 = \log_{24}(54*24^5) = \log_{24}(2^{16}3^{8}) = 8\log_{24}(12)$.
Multiplying together gives $-24\log_{12}(24)\log_{24}(12) = -24\log_{12}(12) = -24$. Adding $25$ to this gives $1$, which is your answer.
A: $ 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
$\frac{1-2X}{X-2}=\frac{1-3Y}{Y-3}$
$(1-2X)(Y-3)=(1-3Y)(X-2)$
$Y-2XY-3+6X=X-3XY-2+6Y$
$Y-2XY-3+6X-X+3XY+2-6Y=0$
$XY+5(X-Y)-1=0$
$XY+5(X-Y)=1$
A: $X = \log_{12} 18$ & $Y= \log_{24} 54$
Let $a=\log_{2} 3$
$\Rightarrow X= {2a+1 \over 2+a}$
$\Rightarrow a= {2X-1 \over 2-X}$
Also,
$Y= {3a+1 \over 3+a}$
$\Rightarrow Y={5X-1 \over 5-X}$
$\Rightarrow XY+5(X-Y)=1$
