I am studying questions for a probability exam. I am stuck on derived distributions. One of my textbook's questions asks: If $X$ is a random variable uniformly distributed between $-1$ and $1$, find the PDF of $-\ln|x|$.
The solution in the textbook is given above: 
I don't understand why they have the initial condition for $y \ge 0$ on the first line. Won't $\ln|X|$ always give a positive number, making $-\ln|X|$ negative, and always mapping $Y$ to a negative number?
Second, I don't understand how they go from $P(X \ge e^{-y}) + P(X \le -e^{-y}) = 1 - e^{-y}$. I tried using the CDF of the exponential distribution to evaluate this, but I didn't get anywhere.
Any help would be greatly appreciated! Thank you.
Firstly, I think you meant the distribution of $X$ to be uniform over $(-1,1)$. So, clearly the absolute value of X will always be a fraction. The logarithm of a fraction is always negative; so, $-ln|X|$ is always a non-negative quantity. Hence $Y = -ln|X|$ is always a non-negative random variable.
Next, if $X$ is uniform over $(-1,1)$, then the cdf of $X$, $$F_X(x) = P[X \le x] = \begin{cases} 0 & \text{for $x \le -1$} \\ (1+x)/2 & \text{for $-1<x<1$} \\ 1 & \text{for $x \ge 1$} \end{cases}$$
Use this to find why $P[X \ge e^{-y}] + P[X \le -e^{-y}] = 1 - e^{-y}$.