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

$$
8^{|x-a|} \log _{1 / 5}\left(x^{2}+2 x+5\right)+2^{x^{2}+2 x} \log _{\sqrt{5}}(3|x-a|+4)=0
$$

have exactly three solutions?
Answer: For $a=-\frac{7}{4},-1,-\frac{1}{4}$.

Solution. Let's simplify the expression:

$$
\begin{gathered}
8^{|x-a|} \log _{1 / 5}\left(x^{2}+2 x+5\right)+2^{x^{2}+2 x} \log _{\sqrt{5}}(3|x-a|+4)=0 \\
2^{3|x-a|}\left(-\log _{5}\left(x^{2}+2 x+5\right)\right)+2^{x^{2}+2 x} 2 \log _{5}(3|x-a|+4)=0 \\
2^{3|x-a|+4}\left(-\log _{5}\left(x^{2}+2 x+5\right)\right)+2^{x^{2}+2 x+5} \log _{5}(3|x-a|+4)=0 \\
\frac{2^{3|x-a|+4}}{\log _{5}(3|x-a|+4)}+\frac{2^{x^{2}+2 x+5}}{\left(-\log _{5}\left(x^{2}+2 x+5\right)\right)}=0
\end{gathered}
$$

Let $3|x-a|+4$ be denoted by $u$, and $x^{2}+2 x+5$ by $v$. Note that $u \geqslant 4, v \geqslant 4$. We will investigate the behavior of the function $f(z)=\frac{2^{2}}{\log _{5} z}$ for $z \geqslant 4$, specifically, we will show that it is monotonic on this ray. To do this, it is sufficient to show that its derivative is sign-constant on it.

$$
f^{\prime}(z)=\frac{(\ln 2) 2^{z} \log _{5} z-2^{z} \frac{1}{\ln 5} \frac{1}{z}}{\left(\log _{5} z\right)^{2}}=\frac{2^{z}}{\left(\log _{5} z\right)^{2}} \cdot\left((\ln 2) \log _{5} z-\frac{1}{(\ln 5) z}\right)
$$

We will show that $\left((\ln 2) \log _{5} z-\frac{1}{(\ln 5) z}\right)>0$ for $z \geqslant 4$. Indeed, $(\ln 2) \log _{5} z \geqslant \ln 2 \log _{5} 4=$ $\frac{\ln 2 \cdot \ln 4}{\ln 5}>\frac{\ln 2}{\ln 8}=\log _{8} 2=\frac{1}{3}>\frac{1}{4} \geqslant \frac{1}{(\ln 5) z}$. Therefore, $f^{\prime}(z)>0$ for $z \geqslant 4$.

The function $f$ is monotonic and $f(u)=f(v)$, hence $u=v$, that is, $3|x-a|+4=x^{2}+2 x+5$, which simplifies to $3|x-a|=(x+1)^{2}$. By plotting the graphs of the functions $g(x)=3|x-a|$ and $h(x)=(x+1)^{2}$, it is easy to see that for there to be exactly three solutions, either their vertices must coincide, or there must be a tangency. The first occurs when $a=-1$, and the second when $a=-1 \pm \frac{3}{4}$.