10. For what value of $a$ does the inequality $\log _{\frac{1}{4}}\left(\sqrt{x^{2}+a x+5}+1\right) \cdot \log _{5}\left(x^{2}+a x+6\right)+\log _{a} 3 \geqslant 0$ have exactly one solution?
10. Clearly $a>0, a \neq 1$. Let $u=x^{2}+a x+5$, the original inequality can be transformed into
$$
\frac{\log _{3}(\sqrt{u}+1)}{\log _{3} \frac{1}{a}} \cdot \log _{5}(u+1)+\frac{1}{\log _{3} a} \geqslant 0 .
$$
(1) When $a \in(0,1)$, the original inequality can be transformed into $\log _{3}(\sqrt{u}+1) \cdot \log _{5}(u+1) \geqslant 1$. Since the functions $y=\log _{3}(\sqrt{u}+1)$ and $y=\log _{5}(u+1)$ are both monotonically increasing functions with a range of $[0,+\infty)$, their product is also a monotonically increasing function. Noting that, when $u=4$, $\log _{3}(\sqrt{4}+1) \cdot \log _{5}(4+1)=1$. Therefore, $u \geqslant 4, x^{2}+a x+5 \geqslant 4$, and this inequality has infinitely many solutions, which is not acceptable.
(2) When $a>1$, the original inequality can be transformed into $\log _{3}(\sqrt{u}+1) \cdot \log _{5}(u+1) \leqslant 1$. Similarly, $0 \leqslant u \leqslant 4, 0 \leqslant x^{2}+a x+5 \leqslant 4$, and this only has a unique solution when $x^{2}+a x+5=4$, thus when $a=2$, this inequality has a unique solution $x=-1$. In summary, the original inequality has a unique solution if and only if $a=2$.