$$y = \ln (|\ln x|)$$
Attempt at solution:
For the function to be defined, $|\ln x|> 0$
$\implies ln x > 0 $
$\implies x > 1$ [$\because$ anti-logging both sides.]
But clearly, the function is valid if $x\in(0,1)$ as well. I am not sure what I am doing wrong. I might be doing something completely illegal here. So sorry about that!
I am looking for a mechanical way of solving these types of problems. So solutions which don't involve graphs (if possible) are requested.
Thanks in advance!
Ps. I have no clue on how to begin calculating for the range.
As you observed, $y=\ln(x)$ is defined for $x>0$. Notice that $|\ln(x)|$ is always greater than $0$, except at $x=1$, where $|\ln(x)|=0$. Therefore, we can take all $x>0$ where $x\neq 1$, and there will be a vertical asymptote at $x=1$ which "splits" the function into two pieces. Together, their domain is $D=\lbrace x>0 |x\neq 1\rbrace$ and the range is $R=\mathbb{R}$. Just looking at the first "piece" of the function with domain $D_1=\lbrace x\in\mathbb{R} |0<x<1\rbrace$, we can clearly see that the range is all reals. This is because the function $\ln(x)$ restricted to the domain $D_1$ has range $R_1=\lbrace x\in\mathbb{R}|-\infty<x<0\rbrace$, so $|\ln(x)|$ has range $R_2=\lbrace x\in\mathbb{R}|0<x<\infty\rbrace$ which is the original domain of the natural log function!