Question

Solve :

$\frac{\log a}{x+y-2z}=\frac{\log b}{y+z-2x}=\frac{\log c}{z+x-2y}$, show that $ab{c}^{4z-2x-2y}=1$ ?

Answer 1

for, $\frac{\log a}{x+y-2z}=\frac{\log b}{y+z-2x}$

$\Rightarrow (y+z-2x)\log a=(x+y-2z)\log b$

$\Rightarrow {\log a}^{(y+z-2x)}={\log b}^{(x+y-2z)}(\because \log a^b=b\log a)$

$\Rightarrow a^{(y+z-2x)}=b^{(x+y-2z)}⟶\left(1\right)$

similarly for, $\frac{\log a}{x+y-2z}=\frac{\log c}{z+x-2y}$

$\Rightarrow a^{(z+x-2y)}=c^{(x+y-2z)}⟶\left(2\right)$

$\left(1\right)\times \left(2\right)$

$\Rightarrow a^{(y+z-2x)}.a^{(z+x-2y)}=b^{(x+y-2z)}.c^{(x+y-2z)}$

$\Rightarrow a^{(2z-x-y)}={\left(bc\right)}^{(x+y-2z)}$

$\Rightarrow {\left(abc\right)}^{(4z-2x-2y)}=1$