How do you approach a logarithm question with absolute value at its base? So I'm a third year high school student and I'm stuck at this question. How do you approach this logarithm to get the domain of x?
$$\log_{|1-x|}(x+5) > 2$$
The answer is $-1 < x < 0$ and $2 < x < 4$
Also, how do you graph the logarithm function with an absolute value? I tried desmos but it doesnt seem to allow adding absolute value in the log base.
Thank you very much.
 A: $$\log_{|1-x|}(x+5) = \frac{\log(x+5)}{\log|1-x|}> 2$$
Firstly $x + 5 > 0$ and $\log|1-x| \ne 0$, that is $x > -5$ and $x \ne 1$
Now if $\log|1-x| > 0$, that is $x < 0 $ or $ x > 2$, you get:
$$\log(x+5) > 2\log|1-x|$$
$$x+5 > (1-x)^2, -1 < x < 4$$
that is: $-1 < x < 0$ or $2 < x < 4$
If $\log|1-x| < 0$, that is $0 < x < 2$, you get:
$$x+5 < (1-x)^2$$
No solution in this case
By consider all above, you get:
$-1 < x < 0$ or $2 < x < 4$
A: For $x>1$ we get $$\frac{\log(x+5)}{\log(x-1)}>2$$
And for $$-5<x<1$$ we get $$\frac{\log(x+5)}{\log(1-x)}>2$$ to solve.
A: We need the base $|1-x|> 0$ and $|1-x|\ne 1$ .
When $0<|1-x|<1$, i.e. $0<x<1$ or $1<x<2$, the logarithm is strictly decreasing and hence
\begin{align*}
x+5&<|1-x|^2\\
x+5&<1-2x+x^2\\
x^2-3x-4&>0\\
x<-1 &\textrm{ or }x>4
\end{align*}
and does not ssatisfy the condition $0<|1-x|<1$. No solution in this case.
When $|1-x|>1$, i.e. $x<0$ or $x>2$, the logarithm is strictly increasing and hence
\begin{align*}
x+5&>|1-x|^2\\
x+5&>1-2x+x^2\\
x^2-3x-4&<0\\
-1<x&<4
\end{align*}
Combining with the condition $|1-x|>1$, we have $-1<x<0$ or $2<x<4$.
A: 
$$f(x) = \log_{|1-x|}(x+5) = \dfrac{\log(x+5)}{\log(|1-x|)}$$
First we note that the domain of $\log(x+5)$ is $(-5, \infty)$.
Since $|1-x|=0$ when x = 1. We must remove $x=1$ from the domain of $f(x)$.
Since $\log |1-x| = 0$ for $x=0$ and $x=2$, we must remove  $x=0$ and $x=2$ from the domain of $f(x)$.
So the domain of $f(x)$ is 
$$(-5, 0) \cup (0,1) \cup (1,2) \cup (2, \infty)$$
Note, we could make $f(x)$ continuous at $x=1$ by defining $f(1)=0$.
Everyone has already done the math for you. But, from the graph, you can see that the solution set is $(-1,0) \cup (2,4)$.
