smc

Notice that all of the graphs $y={\log }_{a}x,$ $y=2{\log }_{a}x,$ and $y=3{\log }_{a}x$ have a domain of $(0,\infty ]$. Also notice that $y={\log }_{a}x$ is the furthest to the right, as adding coefficients in front of the $\log$ part only makes the graph steeper. Since $A$ is on the graph of $y={\log }_{a}x$ and $B$ is on the graph of $y=2{\log }_{a}x$, $A$ must be to the right of $B$. We are told that $\overline{AB}$ is parallel to the x-axis. Let $A$ be the point $(x,y)$. Then the points $B$ and $C$ are $(x-6,y)$ and $(x-6,y+6)$ respectively. Substituting these coordinates into the equations given yields $y={\log }_{a}x,$ $y=2{\log }_{a}x-6,$ and $y+6=3{\log }_{a}x-6$. Rearranging a bit, we get the following equations: $1)a^y=x$ $2)a^y=(x-6{)}^2$ $3)a^{y+6}=(x-6{)}^3$ Using equations 1 and 2, we get $x=(x-6{)}^2$. Solving yields $x=9,x=-4$. However, $-4$ is extraneous so $x=9$ (also because we know $A$ has to have a positive $x$-coordinate). Using the second and third equations, we get $\frac{a^{y+6}}{a^y}=\frac{(x-6{)}^3}{(x-6{)}^2}$ $\Rightarrow$ $a^6=x-6$. Plugging in $9$ for $x$ yields $a^6=3$, or $a=\sqrt[6]{3}\Rightarrow \boxed{\sqrt[6]{3}}$.