Question

If $\frac{\log \left(x\right)}{b-c}=\frac{\log \left(y\right)}{c-a}=\frac{\log \left(z\right)}{a-b}$, then $xyz=?$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B

$1$

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 $\log \left(z\right)=k(a-b)$

Taking antilog on both sides, we get,

$x=e^{k(a-b)},y=e^{k(b-c)}$ and $z=e^{k(c-a)}$ $\left[\because {\log }_b\left(a\right)=c\Rightarrow a=b^c\right]$

Substituting these values in $xyz$, we get

$$\begin{gathered} e^{k(a-b)}·e^{k\left(b-c\right)}.e^{k\left(c-a\right)} \\ =e^{k(a-b+b-c+c-a)}\mathbf{}\left[\because a^m\times a^n=a^{\left(m+n\right)}\right] \\ =e^{k\left(0\right)} \\ =e^0 \\ =1 \end{gathered}$$

Hence, option(B) i.e. $1$, is the correct answer.