Question

If $\frac{\ln x}{b-c}=\frac{\ln y}{c-a}=\frac{\ln z}{a-b}$, then which of the following option(s) is/are true ?

• A. $xyz=1$
• B. $x^a{y}^b{z}^c=1$
• C. $x^{b+c}{y}^{c+a}{z}^{a+b}=1$
• D. $xyz=x^a{y}^b{z}^c$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $xyz=1$
B $x^a{y}^b{z}^c=1$
C $x^{b+c}{y}^{c+a}{z}^{a+b}=1$
D $xyz=x^a{y}^b{z}^c$

Let $\frac{\ln x}{b-c}=\frac{\ln y}{c-a}=\frac{\ln z}{a-b}=k\left(\text{say}\right)$
$\Rightarrow \ln x=k(b-c),\ln y=k(c-b),\ln z=k(a-b)$

$\therefore \ln x+\ln y+\ln z=k(b-c)+k(c-b)+k(a-b)=0$
$\Rightarrow \ln xyz=0$
$\Rightarrow xyz=1$

Also, $x=e^{k(b-c)},y=e^{k(c-a)},z=e^{k(a-b)}$
$\therefore x^a{y}^b{z}^c=e^{ka(b-c)}\cdot e^{kb(c-a)}\cdot e^{kc(a-b)}=e^0=1$

Similarly, $x^{b+c}{y}^{c+a}{z}^{a+b}=e^{k(b^2-c^2)}\cdot e^{k(c^2-a^2)}\cdot e^{k(a^2-b^2)}=e^0=1$