If `log_(theta)x-log_(theta)y= a, log_(theta)y-log_(theta)z=b` and `log_(theta)z- log_(theta)x=c`, then find the value of `(x/y)^(b-c) x(y/z)^(c-a)x(z/x)^(a-b)`

Explanation

`log_(theta)x-log_(theta)y=a rArr log_(theta)x/y=a rArr x/y= e^(a)`
`log_(theta)y-log_(theta)z = b rArr log_(theta)y/z=b rArr y/z=e^(b)`
`log_(theta)z-log_(theta)x=c rArr log_(theta)z/x=c rArr z/x=e^(c )`
`therefore (e^(a))^(b-c) xx (e^(b))^(c-a) xx (e^(c))^(a-b)`
`e^(a(b-c)+b(c-a)+c(a-b)) = e^(0)=1`