# 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 $$({x\over y})^{b-c}$$ $$\times$$ $$({y\over z})^{c-a}$$ $$\times$$ $$({z\over x})^{a-b}$$.

## Solution :

$$log_e x$$ – $$log_e y$$ = a $$\implies$$ $$log_e {x\over y}$$ = a $$\implies$$ $$x\over y$$ = $$e^a$$

$$log_e y$$ – $$log_e z$$ = b $$\implies$$ $$log_e {y\over z}$$ = b $$\implies$$ $$y\over z$$ = $$e^b$$

$$log_e z$$ – $$log_e x$$ = c $$\implies$$ $$log_e {z\over x}$$ = c $$\implies$$ $$z\over x$$ = $$e^c$$

$$\therefore$$  $$(e^a)^{b-c}$$ $$\times$$ $$(e^b)^{c-a}$$ $$\times$$ $$(e^c)^{a-b}$$

= $$e^{a(b-c)+b(c-a)+c(a-b)}$$ = $$e^0$$ = 1

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