Solve : `(log)_((x+3))(x^2-x)<1`

Explanation

`log_(x+3)(x^(2)-x) lt 1`
` x (x-1) gt 0`
` rArr x gt 1 or x lt 0` (i)
If ` x +3 gt 1`
` rArr x gt - 2` (ii)
then `x^(2)- x lt x+ 3`
` or x ^(2) - 2x - 3 lt 0`
` or (x-3)(x+1) lt 0`
Hence, ` x in (-1, 0) cup(1, 3)`[using (i) and (ii)]
If ` 0 lt x+ 3 lt 1`
` :. - 3 lt x lt - 2` (iii)
then ` x^(2) - x gt x+ 3`
` or x^(2) - 2x - 3 gt 0`
` or ( x- 3) (x+ 1) gt 0`
Hence, ` x in (-3, -2)` [using (i) and (iii) ]
Thus, ` x in (-1, 0) cup(1, 3) cup(-3, -2)`