Question

If $\frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}$, then which of the following is 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

$xyz=x^a{y}^b{z}^c$

$\frac{logx}{b-c}=\frac{logy}{c-a}=\frac{logz}{a-b}=k\left(say\right)\Rightarrow logx=k(b-c),logy=k(c-a),logz=k(a-b)\Rightarrow x=e^{k(b-c)},y=e^{k(c-a)},z=e^{k(a-b)}\therefore xyz=e^{k(b-c)+k(c-a)+k(a-b)}=e^0=1x^a{y}^b{z}^c=e^{k(b-c)a+k(c-a)b+k(a-b)c}=e^0=1=xyz{x}^{b+c}{y}^{c+a}{z}^{a+b}=e^{k(b^2-c^2)+k(c^2-a^2)}+k(a^2-b^2)=e^0=1.$
So, All options are correct.