Question

If $\frac {Log\,x}{b-c} = \frac {Log\,y}{c-a}=\frac {Log \,z}{a-b}$ then the value of $x^{b+c} . y^{c+a}.z^{a+b}$ is

• A. 1
• B. 2
• C. 0
• D. -1

Answer 1

Answer: (A)

1

Answer 2

Given, $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say) $\dots(i)$ $\Rightarrow x=e^{k(b-c)}, y=e^{k(c-a)}, z=e^{k(a-b)}$ Now, $x^{b +c} \cdot y^{c +a} \cdot z^{a +b}$ $=e^{k(b-c)(b +c)} \cdot e^{k(c +a)(c-a)} \cdot e^{k(a+ b)(a-b)}$ $=e^{k\left(b^{2}-c^{2}\right)} \cdot e^{k\left(c^{2}-a^{2}\right)} \cdot e^{k\left(a^{2}-b^{2}\right)}$ $=e^{k\left(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}\right)}=e^{k \cdot 0}=1$