Solve `log_2|x-1|<1`

Explanation

`log_2 |x-1| lt 1`

`=>|x-1| lt 2^1`

`=>|x-1| lt 2`

Case 1: When `x ge1`,

Then, `x -1 lt 2`

`=> x lt 3`

`:. x in [1,3)`


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