Equation help

**saha.subham** · January 26th 2011, 01:46 AM

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

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

please tell the steps.

**HallsofIvy** · January 26th 2011, 05:53 AM

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

**Unknown008** · January 26th 2011, 06:52 AM

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

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

?

**saha.subham** · January 28th 2011, 09:06 PM

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

**saha.subham** · January 28th 2011, 09:07 PM

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.

**harish21** · January 28th 2011, 09:43 PM

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

**Unknown008** · January 28th 2011, 09:45 PM

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.

**saha.subham** · January 29th 2011, 02:52 AM

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?

**HallsofIvy** · January 29th 2011, 10:31 AM

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.

**Unknown008** · January 29th 2011, 11:58 PM

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:::

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

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

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

**saha.subham** · January 30th 2011, 02:18 AM

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

**Archie Meade** · January 30th 2011, 04:41 AM

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

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

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

**Unknown008** · January 30th 2011, 04:44 AM

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

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