Square $ABCD$ has area $4,$ and $\overline{AB}$ is parallel to the 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 $a?$

Following the same steps with side length $2$ (area $4$): $t\pm2=\frac{3t}2$, so $t=\pm4$.

1. If $t=4$, let $u=a^2$. Then $|u^2-u|=2$; $u^2-u+2=0$ has no real root, so $u^2-u-2=0$, $(u-2)(u+1)=0$, $u=2$, and $a=\sqrt2$.
2. If $t=-4$, let $u=a^{-2}$. The same equation gives $u=2$, so $a=\frac1{\sqrt2}$. Check: with $a=\frac1{\sqrt2}$, the points $A(4,-4)$, $B(2,-4)$, $C(2,-6)$, $D(4,-6)$ form a square of area $4$, and $\log_a4=-4$, $2\log_a2=-4$, $3\log_a2=-6$.
3. So $a=\boxed{\sqrt2}$ or $a=\boxed{\frac{\sqrt2}2}$.