Question

If $\frac{\log \left(x\right)}{b-c}=\frac{\log \left(y\right)}{c-a}=\frac{\log \left(z\right)}{a-b}$, then the value of $x^{b+c}·y^{c+a}·z^{a+b}$ is:

• A. $1$
• B. $2$
• C. $0$
• D. $-1$

Answer 1

The correct option is A

$1$

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,

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

Substituting these values in $x^{b+c}·y^{c+a}·z^{a+b}$, we get,

$$\begin{gathered} {\left[e^{k(b-c)}\right]}^{{}^{\left(b+c\right)}}·{\left[e^{k(c-a)}\right]}^{{}^{\left(c+a\right)}}·{\left[e^{k(a-b)}\right]}^{{}^{\left(a+b\right)}} \\ =e^{k(b^2-a^2)}·e^{k(c^2-a^2)}·e^{k(a^2-b^2)}\mathbf{[}\mathbf{\because }{\left(a^m\right)}^{\mathbf{n}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{n}}\mathbf{]} \\ =e^{k(b^2-a^2+c^2-a^2+a^2-b^2)}\mathbf{[}\mathbf{\because }{\mathbf{a}}^{\mathbf{m}}\mathbf{·}{\mathbf{a}}^{\mathbf{n}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{+}\mathbf{n}}] \\ =e^{k\left(0\right)} \\ =1 \end{gathered}$$

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