Question

If $\frac{\log \left(x\right)}{b-c}=\frac{\log \left(y\right)}{c-a}=\frac{\log \left(z\right)}{a-b}$, then which of the following is true?

• A. $xyz=1$
• B. $x^a{y}^b{z}^c=1$
• C. $xyz=x^a{y}^b{z}^c$
• D. None of these

Answer 1

The correct option is A

$xyz=1$

Explanation for the correct option:

Given, $\frac{\log \left(x\right)}{b-c}=\frac{\log \left(y\right)}{c-a}=\frac{\log \left(z\right)}{a-b}$

Let this be $\frac{\log \left(x\right)}{b-c}=\frac{\log \left(y\right)}{c-a}=\frac{\log \left(z\right)}{a-b}=k$

So, $\log \left(x\right)=k(b-c),\log \left(y\right)=k(c-a)$ and $\log \left(z\right)=k(a-b)$

Adding these three, we get,

$$\begin{gathered} \log \left(x\right)+\log \left(y\right)+\log \left(z\right)=k(b-c+c-a+a-b) \\ \Rightarrow \log \left(xyz\right)=k\left(0\right)\mathbf{\left[}\mathbf{\because }\mathbf{l}\mathbf{o}\mathbf{g}\left(a\right)\mathbf{+}\mathbf{l}\mathbf{o}\mathbf{g}\left(b\right)\mathbf{=}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}\right] \\ \Rightarrow \log \left(xyz\right)=0 \\ \Rightarrow xyz=e^0\mathbf{}\left[\because lo{g}_b\left(a\right)=c\Rightarrow a=b^c\right] \\ \Rightarrow xyz=1 \end{gathered}$$

Hence, option(A) i.e. $xyz=1$, is the correct answer.