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(log)(0. 5)(3-x)/(x+2)<0...

`(log)_(0. 5)(3-x)/(x+2)<0`

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`log_0.5 ((3-x)/(x+2)) lt 0`
Here, `((3-x)/(x+2))` should be greater than `0` as we can not have logarithms of negative values.
`:. (3-x)/(x+2) gt 0`
`=>x !=-2 and x lt 3`
`=> x in (-2,3) ->(1)`
Now, `log_0.5 ((3-x)/(x+2)) lt 0`
`=> (3-x)/(x+2) lt 0^(0.5)`
`=> (3-x)/(x+2) lt 1`
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