1.

Solve `log_2|x-1|

Answer» `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)`
Case 2: When `x le 1`,
Then, `-x +1 lt 2`
`=> -x lt 1`
`=> x gt -1`
`:. x in (-1,1]`
So, common solution for these two cases will be,
`x in (-1,3).`


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