Solve: `(log)_(x+1/x)(log_2(x-1)/(x+2))>0`

Explanation

` log_(x+1/x)(log_(2).(x-1)/(x+1))gt 0`
` :. (x-1)/(x+1) gt 0`
` rArr x lt -2 or x gt 1`(i)
Also ` x+1/x gt 0`.Therefore,
`(x^(2)+1)/x gt 0`
` rArr x gt 0 `(ii)
From Eqs. (i) and (ii), ` x gt 1`.
Now ` log_(x+1/x)(log_(2). (x-1)/(x+2) ) gt 0 `
`rArr log_(2).(x-1)/(x+2) gt 1" "(.:'x+1/x gt 1" for "x gt 1)`
` or (x-1)/(x+2)gt 2`
` or (x+ 5)/(x+2) lt 0`
Hence, ` x in (-5, -2)`, which is not possible as ` x gt 1`.