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

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You can express all those logs in terms of $\log2$ and $\log3$. – Gerry Myerson Jul 2 '12 at 12:02
@Gerry is correct. Combining everything after the conversion gets a little hairy, but the result is compact. – Blue Jul 2 '12 at 12:16
What is lg supposed to be? – Gerry Myerson Jul 2 '12 at 12:17
lg is same as log @GerryMyerson – Bazinga Jul 2 '12 at 12:28
Is the exclamation point supposed to be a factorial? If not (as it seems from the answer you accepted) please delete it. – Ross Millikan Jul 2 '12 at 13:43

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,

$$I= \frac{\log 3^2\cdot2}{\log 2^2\cdot 3}.\frac{\log 3^3 \cdot 2}{\log 2^3\cdot 3}+5\left(\frac{\log 3^2\cdot2}{\log 2^2\cdot3}-\frac{\log 3^3\cdot2}{\log 2^3\cdot3}\right)= \frac{2x+y}{2y+x}\cdot\frac{3x+y}{3y+x}+5 \left( \frac{2x+y}{2y+x}-\frac{3x+y}{3y+x} \right)$$

$$=\frac{6x^2+5xy+y^2+10x^2+35xy+15y^2-15x^2-35xy-10y^2}{(2y+x)(3y+x)}=\frac{x^2+5xy+6y^2}{x^2+5xy+6y^2}=1$$

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That's what I did. Considering the answer, though, I wonder if there might be a slicker way to get to it. – Blue Jul 2 '12 at 12:29
If you collect terms like $X(Y+5)-5Y$, then it becomes a little easier. – Aang Jul 2 '12 at 12:34

$X = \log_{12} 18$ 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

$\frac{1-2X}{X-2}=\frac{1-3Y}{Y-3}$

$(1-2X)(Y-3)=(1-3Y)(X-2)$

$Y-2XY-3+6X=X-3XY-2+6Y$

$Y-2XY-3+6X-X+3XY+2-6Y=0$

$XY+5(X-Y)-1=0$

$XY+5(X-Y)=1$

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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.

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