8. If the inequality $\log _{\frac{1}{a}}\left(\sqrt{x^{2}+a x+5}+1\right) \cdot \log _{5}\left(x^{2}+a x+6\right)+$ $\log _{3} a \geqslant 0$ has exactly one solution for $x$, then the range of values for $a$ is $\qquad$
8. $a=2$

Analysis: It is easy to know that $a>0, a \neq 1$, let $u=x^{2}+a x+5$, the original inequality can be transformed into
$$
\frac{\log _{3}(\sqrt{u}+1)}{-\log _{3} a} \cdot \log _{5}(u+1)+\frac{1}{\log _{3} a} \geqslant 0
$$
(1) When $01$, the original inequality becomes $\log _{3}(\sqrt{u}+1) \cdot \log _{5}(u+1) \leqslant 1$
Given $f(4)=1$, (2) is equivalent to $0 \leqslant u \leqslant 4$
That is
$$
0 \leqslant x^{2}+a x+5 \leqslant 4
$$

From the above equation, the original inequality has only one solution when $x^{2}+a x+5=4$ has a unique solution (two identical roots). By $\Delta=a^{2}-4=0$ we get $a=2$, at this time the inequality $0 \leqslant x^{2}+2 x+5 \leqslant 4$ can be solved as $x=-1$.
In summary, the original inequality has only one solution when $a=2$.