Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
2  
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
add comment

3 Answers

up vote 1 down vote accepted

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,

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

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

share|improve this answer
    
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
1  
If you collect terms like $X(Y+5)-5Y$, then it becomes a little easier. –  Aang Jul 2 '12 at 12:34
add comment

$ 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

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

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

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

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

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

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.