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Solve `log_2|x-1| |
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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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