Question

If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ then prove that $x^{b+c}.y^{c+a}.z^{b+a}=1$.

Answer 1

Given,

$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$(let). [ Where $k$ is variation constant]

Then $\log x=k(b-c)$

or, $x=e^{k(b-c)}$.....(1) and similarly $y=e^{k(c-a)}$......(2) and $z=e^{k(a-b)}$.....(3).

Now,

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

$=e^{k(b^2-c^2)}.e^{k(c^2-a^2)}.e^{k(a^2-b^2)}$

$=e^{k(b^2-c^2+c^2-a^2+a^2-b^2)}$

$=e^0$

$=1$.