Question

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

Answer 1

Given, $\frac{\mathrm{logx}}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b},=k$ (Let)

Then,

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

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

Now,

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

$=e^{k(b^2-c^2)}\times e^{k(c^2-a^2)}\times e^{k(a^2-b^2)}$ [ Using (1)]

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

$=e^0$

$=1$.