5. Given that $[x]$ represents the greatest integer not exceeding $x$. Then the number of solutions to the equation
$$
3^{2 x}-\left[10 \times 3^{x+1}\right]+\sqrt{3^{2 x}-10 \times 3^{x+1}+82}=-80
$$
is
5.2.

The original equation can be transformed into
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1\right)=0 . \\
\therefore 3^{2 x}-\left[10 \times 3^{x+1}\right]+81=0,
\end{array}
$$

It is only necessary that $3^{2 x}-10 \times 3^{x+1}+81 \leqslant 0$,
which means $\left(3^{x}-3\right)\left(3^{x}-27\right) \leqslant 0 \Rightarrow 3 \leqslant 3^{x} \leqslant 27$.
$$
\therefore 1 \leqslant x \leqslant 3 \text {. }
$$

When $x=1,3$, the original equation holds. When $x=2$, the original equation has no solution. Therefore, the original equation has two solutions.