Problem 7. For what values of the parameter $a$ does the equation

$$
a \log _{3}^{2} x-(2 a+3) \log _{3} x+6=0
$$

have exactly two roots, one of which is three times the other?
Answer: 1,3.

Solution. Let $t=\log _{3} x$, then the equation takes the form $a t^{2}-(2 a+3) t+6=0$ or $t^{2}-\frac{2 a+3}{a} t+\frac{6}{a}=$ $0(a \neq 0)$. Note that $2 \cdot \frac{3}{a}=\frac{6}{a}, 2+\frac{3}{a}=\frac{2 a+3}{a}$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -2 and $\frac{3}{a}$. We make the reverse substitution: $\log _{3} x=2$ or $\log _{3} x=\frac{3}{a}$, i.e., either $x=9$, or $x=3^{\frac{3}{a}}$. We get two cases: $3 \cdot 9=3^{\frac{3}{a}}$ or $9=3 \cdot 3^{\frac{3}{a}}$. In the first case $a=1$, in the second $-a=3$.