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

$$
a \log _{5}^{2} x-(2 a+5) \log _{5} x+10=0
$$

have exactly two roots, one of which is five times larger than the other?
Answer: $\frac{5}{3}, 5$.

Solution. Let $t=\log _{5} x$, then the equation takes the form $a t^{2}-(2 a+5) t+10=0$. Note that $2 \cdot \frac{5}{a}=\frac{10}{a}, 2+\frac{5}{a}=\frac{2 a+5}{a}$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -2 and $\frac{5}{a}$. We make the reverse substitution: $\log _{5} x=2$ or $\log _{5} x=\frac{5}{a}$, i.e., either $x=25$, or $x=5 \frac{5}{a}$. We get two cases: $5 \cdot 25=5^{\frac{5}{a}}$ or $25=5 \cdot 5^{\frac{5}{a}}$ In the first case $a=\frac{5}{3}$, in the second $-a=5$.