1. ## Equation help

if log x / a-b = log y / b-c = log z / c-a then xyz = ?????

answer - a)2 , b)-1 ,c) 1 ,d) 0

2. Have you read the "Sticky: List of rules used to moderate MHF - please read carefully before posting."?

3. Also, making use of brackets to clear any confusion is important. For example, is your question:

$\dfrac{\log\ x}{a-b} = \dfrac{\log\ y}{b-c} = \dfrac{\log\ z}{c-a}$

or

$\log\ \dfrac{x}{a-b} = \log\ \dfrac{y}{b-c} = \log\ \dfrac{z}{c-a}$

?

4. Originally Posted by Unknown008
Also, making use of brackets to clear any confusion is important. For example, is your question:

$\dfrac{\log\ x}{a-b} = \dfrac{\log\ y}{b-c} = \dfrac{\log\ z}{c-a}$

or

$\log\ \dfrac{x}{a-b} = \log\ \dfrac{y}{b-c} = \log\ \dfrac{z}{c-a}$

?
the top one

5. Originally Posted by HallsofIvy
Have you read the "Sticky: List of rules used to moderate MHF - please read carefully before posting."?
srry whats the problem? i thik this is fro algebra right? plzz guide me.

6. you have to show the steps where you are stuck instead of asking for the entire solution... Thats what HallsofIvy meant..

7. Originally Posted by saha.subham
srry whats the problem? i thik this is fro algebra right? plzz guide me.
The problem is that you have to show what you have tried so far, to solve this problem.

8. Originally Posted by Unknown008
The problem is that you have to show what you have tried so far, to solve this problem.
(x)/a-b = (y )/ b-c and (y)/b-c = (z)/c - a (after cancelling out log.)
now what?

9. Now, you have (b- c)x= (a- b)y and (c- a)y= (b- c)z which you can rewrite as bx- cx= ay- by and cy- ay= bz- cz. Play with those.

10. Hm... actually, you cannot cancel log like this. If it were, because you have a denominator outside the log function. If you want to cancel log, you will have to do it like this:

$\dfrac{1}{a-b} \log (x) = \log x^{\left(\frac{1}{a-b}\right)}$

Then you get:

$x^{\left(\frac{1}{a-b}\right)} = y^{\left(\frac{1}{b-c}\right)} = z^{\left(\frac{1}{c-a}\right)}$

But I don't know if that will work...

I tried using this:

$\log(x) = \dfrac{(a-b) \log (y)}{(b-c)}$

$\log(z) = \dfrac{(c-a) \log (y)}{(b-c)}$

Then, we get:

$\log (x) + \log(y) + \log(z) = \dfrac{(a-b) \log (y)}{(b-c)} + \log(y) + \dfrac{(c-a) \log (y)}{(b-c)}$

Hence,

$\log(xyz) = \dfrac{(a-b) \log (y) + (b-c)\log(y) + (c-a)\log(y)}{(b-c)}$

$\log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}$

$\log(xyz) = \dfrac{\log (y)}{(b-c)}$

Though I'm not too sure where to go from here.

EDIT: From the before last line:::

$\log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}$

$\log(xyz) = \dfrac{[0]\log (y)}{(b-c)}$

$\log(xyz) = \dfrac{0}{(b-c)} = 0$

11. Originally Posted by HallsofIvy
Now, you have (b- c)x= (a- b)y and (c- a)y= (b- c)z which you can rewrite as bx- cx= ay- by and cy- ay= bz- cz. Play with those.
i am not being able to find the value of xyz plzz help

12. If you want to find $xyz$

then you can obtain it from $logx+logy+logz=log(xyz)$

using the properties of logs.

You need to get an expression for the sum of logs.
Hence

$\frac{logx}{a-b}=\frac{logy}{b-c}\Rightarrow\ logy=\left(\frac{b-c}{a-b}\right)logx$

Also

$\frac{logx}{a-b}=\frac{logz}{c-a}\Rightarrow\ logz=\left(\frac{c-a}{a-b}\right)logx$

Therefore

$logx+logy+logz=logx\left(1+\frac{b-c}{a-b}+\frac{c-a}{a-b}\right)=logx\left(\frac{a-b+b-c+c-a}{a-b}\right)=logx\left(\frac{0}{a-b}\right)=0$

$logx+logy+logz=log(xyz)=0\Rightarrow\ xyz=\;?$

13. Oh, I was close but didn't see it... my bad