There are two simultaneous equations. So, we will try to solve the easier one first.
1−2x3−|x−1|=1
⇒1−2x=3−|x−1|
⇒|x−1|=2(1+x)
If x≥1, we get
x−1=2(1+x)⇒x=−3
Not possible
If x<1, we get
1−x=2(1+x)⇒x=−13
Now, putting x=−13, we get
(log{x})2−3log[x]+2=0⇒(log{−13})2−3log[−13]+2=0
Which is not defined as [−13]=0
Hence, no solution exists.