Question

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

Answer 1

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

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

$x={10}^{k(b-c)}$

$y={10}^{k(c-a)}$

$z={10}^{k(a-b)}$

$x^{b+c}{y}^{c+a}{z}^{a+b}={10}^{k(b-c)(b+c)}\times {10}^{k(c-a)(c+a)}\times {10}^{k(a-b)(a+b)}$

$={10}^{k(b^2-c^2+c^2-a^2+a^2-b^2)}$

$={10}^0$

$=1$