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

$$
\log _{4}^{2} x+(a-4) \log _{4} x+a-5=0
$$

have exactly two roots, one of which is four times the other?
Answer: 7.5.

Solution. Let $t=\log _{4} x$, then the equation takes the form $t^{2}+(a-4) t+a-5=0$. Notice that $(-1) \cdot 5-a=a-5,-1+(5-a)=4-a$, from which, by the theorem converse to Vieta's theorem, the roots of this equation are -1 and $5-a$. We make the reverse substitution: $\log _{4} x=-1$ or $\log _{4} x=5-a$, i.e., either $x=\frac{1}{4}$, or $x=4^{5-a}$. We get two cases: $\frac{1}{4}=4 \cdot 4^{5-a}$ or $4 \cdot \frac{1}{4}=4^{5-a}$. In the first case $a=7$, in the second $a=5$.