Square $ABCD$ has area $36,$ and $\overline{AB}$ is [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) to the [x-axis](https://artofproblemsolving.com/wiki/index.php/X-axis). Vertices $A,$ $B$, and $C$ are on the graphs of $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x,$ respectively. What is $a?$
$\mathrm{(A)}\ \sqrt [6]{3}\qquad \mathrm{(B)}\ \sqrt {3}\qquad \mathrm{(C)}\ \sqrt [3]{6}\qquad \mathrm{(D)}\ \sqrt {6}\qquad \mathrm{(E)}\ 6$
Let $x$ be the x-coordinate of $B$ and $C$, and $x_2$ be the x-coordinate of $A$ and $y$ be the y-coordinate of $A$ and $B$. Then $2\log_ax= y \Longrightarrow a^{y/2} = x$ and $\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2$. Since the distance between $A$ and $B$ is $6$, we have $x^2 - x - 6 = 0$, yielding $x = -2, 3$.
However, we can discard the negative root (all three [logarithmic](https://artofproblemsolving.com/wiki/index.php/Logarithm) equations are underneath the line $y = 3$ and above $y = 0$ when $x$ is negative, hence we can't squeeze in a square of side 6). Thus $x = 3$.
Substituting back, $3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x$, so $a = \sqrt[6]{3}\ \ \mathrm{(A)}$.