Question: Just wondering why we have equations defined in the Cartesian coordinate for circles and ellipses: wouldn't graphing those shapes contradict the fundamental property of a function (i.e.: to each element in the domain corresponds a single element in the codomain)?

I know I can *algebraically* restate the equations as a function of y, but that would leave me with a square root whose domain I don't know, so the problem remains.

Answer: 
you are asked to find the probability distribution of the random variable $Y$ that is related to the random variable $X$ by the relation $Y=-2\ln X$, being $X$ uniform in the interval $[0,1]$. You can solve it considering a change of variable applied to the cumulative function $F(x)$:
$$F(x)=\int\_0^x f(x)dx = \int\_\infty^y f(y)\Big|\frac{dx}{dy}\Big|^{-1}dy$$
so that
$$f(y) = f(x)\Big|\frac{dx}{dy}\Big|\_y$$
and in your situation $x=\exp(-y/2)$, so $\Big|\frac{dx}{dy}\Big|\_y = \frac{1}{2}\exp(-y/2)$, so $$f(y)= \frac{1}{2}\exp(-y/2)$$

If you try to integrate $f(y)$ between $\infty$ and $0$ you can verify that it gives you $1$.
$ P(-2logX\le x)=P(logX\ge -x/2)=P(X\ge e^{\frac{-x}{2}})=\int\_{e^{\frac{-x}{2}}}^1dt=1-e^{\frac{-x}{2}}$

$F\_X(x)=1-e^{\frac{-x}{2}}$

hence $f\_X(x)=\frac{1}{2}e^{\frac{-x}{2}}$  

which is the pdf of exponential with parameter $\frac{1}{2}$