How to solve this logarithm inequality with absolute value as its base? How to deal with this ?
$$\log_{|1 - x|} (x+5)>2 $$
the $|1-x|$ is the base of the logarithm.
I tried this below approach but it seems not the complete solution.
\begin{align}
\frac{\log(x+5)}{\log|1-x|} & > 2\\
{\log(x+5)} & > 2{\log|1-x|}\\
{\log(x+5)} & > {\log|1-x|^2}\\
(x+5) & >|1-x|^2\\
(x+5) & >(1-x)^2\\
(x+5) & >1-2x+x^2\\
x^2-3x-4& < 0
\end{align}
$$ -1<x<4 $$
I also checked with wolframalpha.
https://www.wolframalpha.com/input/?i=log+%5Babsolut(1-x),+(x%2B5)%5D%3C2
I appreciate your help.
 A: 

Look at the pictures. You will see there are two cases to the problem. 
Here is a link to : Wolfram Alpha
Hope this helps. If you need help in solving the individual inequalities, comment and I will show this as well. 
A: Hint: Note that the base should be $$\left| 1-x \right| \neq 1\Rightarrow x\neq 0\quad \& \quad x\neq 2$$ and $$x+5>0\quad \Rightarrow \quad x>-5$$ 
A: Developping kingW3's idea, we'll know why that's empty region in the solution.  We first note that $x \ne 1$.

The problem in the question body is the transition from
$$\frac{\log(x+5)}{\log|1-x|} > 2\tag1\label1$$
to
$${\log(x+5)} > 2{\log|1-x|}.\tag2\label2$$
By neglecting case 2 below, we failed to capture the empty region $0 < x < 2$ in the range for $x$.
Case 1: $\log|1-x|>0 \iff |1-x|>1 \iff x<0$ or $x>2$.  Using OP's calculations above, we get $-1 < x < 0$ or $2 < x < 4$, which agrees with the graph of the linked Wolfram Alpha page.
Case 2: $\log|1-x|<0 \iff 0<|1-x|<1 \iff 0 < x < 1$ or $1 < x < 2$.
We should reverse the inequality \eqref{2} in this case since we're multiplying \eqref{1} by a negative denominator $\log|1-x|$.
$${\log(x+5)} < 2{\log|1-x|}$$
As a result, the inequalities in the question body that follow from \eqref{2} should be reversed, until $x^2-3x-4>0$, from which we deduce $x<-1$ or $x>4$.  The intersection of $(-\infty,-1)\cup(4,\infty)$ with $(0,1)\cup(1,2)$ is empty, so no real value of $x$ satisfies case 2.
Hence the solution is $-1<x<0$ or $2<x<4$.
