Question

If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$. show that
$\left(i\right)$ $xyz=1$
$\left(ii\right)$ $x^a{y}^b{z}^c=1$
$\left(iii\right)$ $x^{b+c}{y}^{c+a}{z}^{a+b}=1$

Answer 1

(i)

Here,

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

$\log x=K(b-c)$ -----(1)

$\log y=K(c-a)$------(2)

$\log z=K(a-b)$ --------(3)

add eqns (1) (2) and (3)

$\log x+\log y+\log z=K(b-c+c-a+a-b)$

$\log xyz=K.0=0$

$xyz=1$

(ii)

now,

$\log x=K(b-c)$

$a\log x=Ka(b-c)$

$\log x^a=K(ab-ac)$

similarly ,

$b\log y=Kb(c-a)=K(bc-ab)$

$\log y^a=K(bc-ab)$

$c\log z=K(ac-bc)$

$\log z^c=K(ac-bc)$

add all terms,

$\log x^a+\log y^b+\log z^c=K(ab-ac+bc-ab+ac-bc)=0$

$\log \left(x^a\right)\left(y^b\right)\left(z^c\right)=0$

$\left(x^a\right)\left(y^b\right)\left(z^c\right)=1$

(iii)

$\log x=K(b-c)$

$(b+c)\log x=K(b-c)(b+c)=K(b²-c²)$

similarly,

$\log y=K(c-a)$

$(c+a)\log y=K(c^2-a^2)$

$\log z=K(a-b)$

$(a+b)\log z=K(a^2-b^2)$

add all terms ,

$\log \left(x^{b+c}\right).\left(y^{c+a}\right).\left(z^{a+b}\right)=K(b^2-c^2+c^2-a^2+a^2-b^2)=K\times 0=0$

$x^{b+c}.y^{c+a}.z^{a+b}=1$