Solve `x^(2)+2x-3`

Answer

`x epsilon (-1-sqrt(5),-3)uu(sqrt(5)-1,5)`

Explanation

We have `log_(x^(2)+2x3-)((|x+4|-|x|)/(x-1))gt0`
The given inequation is valid for
`(|x+4|-|x|)/((x-1))gt0`
and `x^(2)+2x-3gt0,!=1`…….i
Now consider the following cases:
Case I If `0ltx^(2)+2x-3lt1`
`implies4lt(x+1)^(2)lt5`
`implies-sqrt(5)lt(x+1)lt-2` or `2ltx+1ltsqrt(5)`
`implies-sqrt(5)-1ltxle-3` or `1ltxltsqrt(5)-1`
`:.x epsilon (-sqrt(5)-1,-3)uu(1,sqrt(5)-1)`...ii
Ten `(|x+4|-|x|)/((x-1))lt1`
Now `xlt-4`, ten `(-(x+4)+x)/((x-1))lt1`
`implies 1+4/(x-1)gt0`
`implies((x+3))/((x-1))gt0`
`:.x epsilon (-oo,-3)uu(1,oo)`
`impliesx epsilon (-oo,-4)[:' x lt -4]`.iii
`-4lexlt0`, then `(x+4+x)/((x-1))-1lt0`
`implies((x+5))/((x-1))lt0`
`:.x epsilon (-5,1)`
`impliesx epsilon [-4,0)[:'-4lexlt0]` ..iv
and `xge0` then `((x+4-x)/((x-1))lt1`
`implies1=4/(x-1)gt0`
`implies((x-5))/((x-1))gt0`
`:.x epsilon (-oo,1)uu(5,oo)`
`impliesx epsilon [0,1)uu(5,oo)[ :' x ge0]`.......v
From eqs (iii), (iv) and (v) we get
`x epsilon (-oo,1)uu(5,oo)` .vi
Now common values in Eq. (ii) and (iv) is
`x epsilon (-sqrt(5)-1,-3)`..vii
Case II If `x^(2)+2x-3gt1`
`impliesx^(2)+2x+1gt5implies(x+1)^(2)gt5`
`impliesx+1lt-sqrt(5)`
or `x+1gtsqrt(5)`
`:.x epsilon(-oo,-1-sqrt(5))uu(sqrt(5)-1,oo)`........viii
Then `(|x+4|-|x|)/((x-1))gt1`
Now `xlt-4`, then `(-4)/(x-1)gt1`
`implies1+4/(x-1)lt0`
`implies (x+3)/(x-1)lt0`
`:. x epsilon (-3,1)`
Which is false `[:' x lt-4]`
`-4lexlt0` then `(2x+4)/((x-1))-1gt0` ltbRgt `implies((x+5))/((x-1))gt0`
`:.x epsilonn (-oo,-5)uu(1,oo)`
Which is false `[:'-4lexlt0]`
and `xge0` then `4/(x-1)gt1`
`implies1-4/(x-1)lt0`
`implies(x-5)/(x-1)lt0`
`:.x epsilon (1,5)`....ix
which is false `[:'xge0]`
Now comon values in Eq (viii) and (ix) is
`:. x epsilon (sqrt(5)-1,5)` .......x
Combinin Eqs (viii) and (x) we get
`x epsilon (-sqrt(5)-1,-3)uu(sqrt(5)-1,5)`