Question

If ${\log }_{e}x-{\log }_{e}y=a,{\log }_{e}y-{\log }_{e}z=b,\&{\log }_{e}z-{\log }_{e}x=c$, then find the value of${\left(\begin{array}{c}\frac{x}{y}\end{array}\right)}^{b-c}\times {\left(\begin{array}{c}\frac{y}{z}\end{array}\right)}^{c-a}\times {\left(\begin{array}{c}\frac{z}{x}\end{array}\right)}^{a-b}$

Answer 1

${\log }_{e}x-{\log }_{e}y=a\Rightarrow {\log }_e\frac{x}{y}=a\Rightarrow \frac{x}{y}=e^a$
${\log }_{e}y-{\log }_{e}z=b\Rightarrow {\log }_e\frac{y}{z}=b\Rightarrow \frac{y}{z}=e^b$
${\log }_{e}z-{\log }_{e}x=c\Rightarrow {\log }_e\frac{z}{x}=c\Rightarrow \frac{z}{x}=e^c$
$\therefore {\left(\begin{array}{c}\frac{x}{y}\end{array}\right)}^{b-c}\times {\left(\begin{array}{c}\frac{y}{z}\end{array}\right)}^{c-a}\times {\left(\begin{array}{c}\frac{z}{x}\end{array}\right)}^{a-b}={\left(\begin{array}{c}{e}^a\end{array}\right)}^{b-c}\times {\left(\begin{array}{c}{e}^b\end{array}\right)}^{c-a}\times {\left(\begin{array}{c}{e}^c\end{array}\right)}^{a-b}=e^0=1$