## Task A-3.3.

Square $A B C D$ with an area of 36 is placed in the coordinate plane such that side $\overline{A B}$ is parallel to the $y$-axis, and points $A, B$, and $C$ lie on the graphs of the functions $f(x)=3 \log _{a} x$, $g(x)=2 \log _{a} x$, and $h(x)=\log _{a} x$, respectively. Determine the number $a$.
## Solution.

Let's denote the coordinates of the points as $A\left(x_{A}, 3 \log _{a} x_{A}\right)$, $B\left(x_{B}, 2 \log _{a} x_{B}\right)$, and $C\left(x_{C}, \log _{a} x_{C}\right)$. Since the area of the square is 36, the length of the side is 6. The side $\overline{A B}$ is parallel to the $y$-axis, so $x_{A}=x_{B}$ and $\left|3 \log _{a} x_{A}-2 \log _{a} x_{B}\right|=6$, which implies $\left|\log _{a} x_{B}\right|=6$.

Furthermore, $2 \log _{a} x_{B}=\log _{a} x_{C}$, i.e., $x_{C}=x_{B}^{2}$.

The side $\overline{B C}$ is parallel to the $x$-axis, so $\left|x_{C}-x_{B}\right|=6$.

Therefore, $x_{B}$ and $x_{C}$ must be greater than 1. We consider the cases $a>1$ and $a<1$.

If $a>1$, then $\log _{a} x_{B}=6$ and $x_{C}=x_{B}+6$, leading to the quadratic equation $x_{B}^{2}-x_{B}-6=0$. The solutions to this equation are $x_{B}=3$ and $x_{B}=-2$, but since $x_{B}$ must be positive, the only possible solution is $x_{B}=3$.

Since $\log _{a} x_{B}=6$, i.e., $x_{B}=a^{6}$, it follows that $a=\sqrt[6]{3}$.

The problem has two solutions, where the results are reciprocals of each other, and the graphs are mirror-symmetric with respect to the $x$-axis. For $a<1$, with this comment, 6 points should be awarded.