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