Question

If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$

Prove:
(i) $xyz=1$
(ii) $x^a{y}^b{z}^c=1$
(iii) $x^{b+c}{y}^{c+a}{z}^{a+b}=1$.

Answer 1

Let $\frac{\log x}{b-c}=\frac{\log y}{c-a}=$ $\frac{\log z}{a-b}=k$.

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

$1.$ $\log \left(xyz\right)=\log x+\log y+\log z=k(b-c+c-a+a-b)$.

That gives us $\log \left(xyz\right)=0$ or $xyz=1$

$2.$ $a.\log x+b.\log y+c.\log z=k(ab-ac+bc-ba+ca-cb)$.

This implies $\log \left(x^a{y}^b{z}^c\right)=0$ or $x^a{y}^b{z}^c=1$.

$3.$ $(b+c).\log x+(c+a).\log y+(a+b).\log z=k(b^2-c^2+c^2-a^2+a^2-b^2)$.

This implies $\log \left(x^{b+c}{y}^{c+a}{z}^{a+b}\right)=0$ or $x^{b+c}{y}^{c+a}{z}^{a+b}=1$.

Hence proved