Problem 8. For what values of the parameter $a$ does the equation

$$
4^{|x-a|} \log _{1 / 3}\left(x^{2}-2 x+4\right)+2^{x^{2}-2 x} \log _{\sqrt{3}}(2|x-a|+3)=0
$$

have exactly three solutions?
Answer: For $a=\frac{1}{2} ; 1 ; \frac{3}{2}$.

Solution. Let's simplify the expression:

$$
4^{|x-a|} \log _{1 / 3}\left(x^{2}-2 x+4\right)+2^{x^{2}-2 x} \log _{\sqrt{3}}(2|x-a|+3)=0
$$

$$
\begin{gathered}
2^{2|x-a|}\left(-\log _{3}\left(x^{2}-2 x+4\right)\right)+2^{x^{2}-2 x} 2 \log _{3}(2|x-a|+3)=0 \\
2^{2|x-a|+3}\left(-\log _{3}\left(x^{2}-2 x+4\right)\right)+2^{x^{2}-2 x+4} \log _{3}(2|x-a|+3)=0 \\
\frac{2^{2|x-a|+3}}{\log _{3}(2|x-a|+3)}+\frac{2^{x^{2}-2 x+4}}{\left(-\log _{3}\left(x^{2}-2 x+4\right)\right)}=0
\end{gathered}
$$

Let $2|x-a|+3$ be denoted by $u$, and $x^{2}-2 x+4$ by $v$. Note that $u \geqslant 3, v \geqslant 3$. We will investigate the behavior of the function $f(z)=\frac{2^{z}}{\log _{3} z}$ for $z \geqslant 3$, specifically showing that it is monotonic on this ray. To do this, it is sufficient to show that its derivative is sign-constant on it.

$$
f^{\prime}(z)=\frac{(\ln 2) 2^{z} \log _{3} z-2^{z} \frac{1}{\ln 3} \frac{1}{z}}{\left(\log _{3} z\right)^{2}}=\frac{2^{z}}{\left(\log _{3} z\right)^{2}} \cdot\left((\ln 2) \log _{3} z-\frac{1}{(\ln 3) z}\right)
$$

We will show that $\left((\ln 2) \log _{3} z-\frac{1}{(\ln 3) z}\right)>0$ for $z \geqslant 3$. Indeed, $(\ln 2) \log _{3} z \geqslant \ln 2 \log _{3} 3=\ln 2>\frac{\ln 2}{\ln 4}=\log _{4} 2=\frac{1}{2}>\frac{1}{3} \geqslant \frac{1}{(\ln 3) z}$. Therefore, $f^{\prime}(z)>0$ for $z \geqslant 3$.

The function $f$ is monotonic and $f(u)=f(v)$, so $u=v$, that is, $2|x-a|+3=x^{2}-2 x+4$, or $2|x-a|=(x-1)^{2}$. By plotting the graphs of the functions $g(x)=2|x-a|$ and $h(x)=(x-1)^{2}$, it is easy to see that to have exactly three solutions, either their vertices must coincide, or there must be a tangency. The first occurs at $a=1$, the second at $a=1 \pm \frac{1}{2}$.