If ` a = 1 + log_(x) yz, b = 1 + log_(y) zx and c = 1 + log xy ` where x, y, z are positive real numbers
different unity , them prove that abc = ab + bc + ca

Options

A. `(1)/(x+1) + (1)/(y+1) + (1)/(z+1) =1`
B. `(1)/(x-1) + (1)/(y-1) + (1)/(z-1) =1`
C. xyz =x + y + z + 1
D. xyz = 1

Answer

A

Explanation

We have,
`x = "log"_(a)bc, y = "log"_(b)ca, z = "log"_(c) ab`
`rArr x + 1 = "log"_(a)bc + "log"_(a) a, y+1 = "log"_(b)ca + "log"_(b)b, z+1 = "log"_(c)ab + "log"_(c)c`
`rArr x+ 1 = "log"_(a)abc, y + 1 = "log"_(b)abc, z+1 = "log"_(c)abc`
`rArr (1)/(x+1) + (1)/(y + 1) + (1)/(z+1) = "log"_(abc)a+"log"_(abc)b + "log"_(abc)c`
`rArr (1)/(x + 1) + (1)/(y + 1) + (1)/(z+1) = "log"_(abc) abc = 1`