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.
Also, making use of brackets to clear any confusion is important. For example, is your question:
$\displaystyle \dfrac{\log\ x}{a-b} = \dfrac{\log\ y}{b-c} = \dfrac{\log\ z}{c-a}$
or
$\displaystyle \log\ \dfrac{x}{a-b} = \log\ \dfrac{y}{b-c} = \log\ \dfrac{z}{c-a}$
?
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:
$\displaystyle \dfrac{1}{a-b} \log (x) = \log x^{\left(\frac{1}{a-b}\right)}$
Then you get:
$\displaystyle x^{\left(\frac{1}{a-b}\right)} = y^{\left(\frac{1}{b-c}\right)} = z^{\left(\frac{1}{c-a}\right)}$
But I don't know if that will work...
I tried using this:
$\displaystyle \log(x) = \dfrac{(a-b) \log (y)}{(b-c)}$
$\displaystyle \log(z) = \dfrac{(c-a) \log (y)}{(b-c)}$
Then, we get:
$\displaystyle \log (x) + \log(y) + \log(z) = \dfrac{(a-b) \log (y)}{(b-c)} + \log(y) + \dfrac{(c-a) \log (y)}{(b-c)}$
Hence,
$\displaystyle \log(xyz) = \dfrac{(a-b) \log (y) + (b-c)\log(y) + (c-a)\log(y)}{(b-c)}$
$\displaystyle \log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}$
$\displaystyle \log(xyz) = \dfrac{\log (y)}{(b-c)}$
Though I'm not too sure where to go from here.
EDIT: From the before last line:::
$\displaystyle \log(xyz) = \dfrac{[(a-b) + (b-c) + (c-a)]\log (y)}{(b-c)}$
$\displaystyle \log(xyz) = \dfrac{[0]\log (y)}{(b-c)}$
$\displaystyle \log(xyz) = \dfrac{0}{(b-c)} = 0$
If you want to find $\displaystyle xyz$
then you can obtain it from $\displaystyle logx+logy+logz=log(xyz)$
using the properties of logs.
You need to get an expression for the sum of logs.
Hence
$\displaystyle \frac{logx}{a-b}=\frac{logy}{b-c}\Rightarrow\ logy=\left(\frac{b-c}{a-b}\right)logx$
Also
$\displaystyle \frac{logx}{a-b}=\frac{logz}{c-a}\Rightarrow\ logz=\left(\frac{c-a}{a-b}\right)logx$
Therefore
$\displaystyle logx+logy+logz=logx\left(1+\frac{b-c}{a-b}+\frac{c-a}{a-b}\right)=logx\left(\frac{a-b+b-c+c-a}{a-b}\right)=logx\left(\frac{0}{a-b}\right)=0$
$\displaystyle logx+logy+logz=log(xyz)=0\Rightarrow\ xyz=\;?$