Solving $|x-1|^{\log_2(4-x)} \le|x-1|^{\log_2 (1+x)}$ Let us solve
$|x-1|^{\log_2(4-x)}\le|x-1|^{\log_2 (1+x)}......(*)$
Let $4-x>0 ~\& ~1+x>0$.......(1)
Case 1: $|x-1|\le 1\implies 0\le x\le 2$........(2)
We get $\log_2(4-x) \color{red}{\ge} \log_2(1+x)\implies 4-x \ge 1+x \implies x\le 3/2$....(3)
The overlap of (1,2,3) gives $x\in [0,3/2]$........(4)
Case 2: $|x-1|\ge 1 \implies x\le 0 ~or~ x\ge 2$....(5)
We get $\log_2(4-x) \color{red}{\le} \log_2(1+x)\implies 4-x <1+x \implies x\ge 3/2$....(6)
Taking overlap of (1,5,6), we get $x \in [2,4)$.
So the final solution is : $[0,3/2] \cup [2,4).$
Now the question is  whether this solution is complete and how else this (*) could be solved?

$|x-1|^{\log_2(4-x)}\le|x-1|^{\log_2 (1+x)}......(1)$
Note that $x=0,1,2$ are already  roots.
Taking $\log_a$ both side ( $a\in (0,1)$ or $a\in(1,\infty))$, let us take $a=2$. We get
$$\log_2(4-x) \log_2|x-1|\color{red}{\le} \log_2(1+x)\log_2|x-1|\quad(2)$$
Since, $\log_2(4-x)$ and $\log_2(1+x)$ have to be real we declare
$$(4-x)>0~\&~ (1+x)>0\quad(3)$$
Case 1: $|x-1|<1\implies 0<x< 2, \log_2|x-1|\color{red}{<}0\quad(4)$,
From (2) we get
$$\log_2(4-x)  \color{red}{\ge}  \log_2(1+x) \implies 4-x \ge 1+x \implies x \le 3/2\quad(5)$$
Taking intersection of (3,4,5) and inclusion of $x=0,1$; we get the solution as
$x\in [0,3/2]\quad (6)$
Case 2: $|x-1|> 1 \implies x<0 ~or~x > 2, \log_2|x-1|\color{red}{>}0 \quad (7)$
This time from (2), we get
$$\log_2(4-x) \color{red}{\le} \log_2(1+x) \implies 4-x \le 1+x \implies x\ge 3/2 \quad (8)$$
Finally, intersection of (3,7,8) and inclusion of $x=2$, gives the solution as $2\le x< 4 \quad (9)$
So the final solution lies in $[0,3/2] \cup [2,4)$.
Note that by choosing the base e.g., $1/2$ wont change the final answer.
