Help on sketching on log functions with absolute

[duhherro]

Also, how would you determine the 2 hybrid functions of

y = |-loge (1-x) - 2|
----> y = |loge (1-x) - 2|



for something like y = |loge (x-2)|
I did, loge (x-2) < 0 and loge (x-2) > 0, and find the x-int of loge (x-2) is x=3.

So you get:
y= -loge (x-2), 2<x<3     ------- so is it possible to use a "rule" for the domain such as x-asymtote < x < x-int. ?
 and loge (x-2), x>3        ------- and hence x > x-int ?