The solution set of ${\log }_8\left(x^2-4x+3\right)\le 1$ is

$\begin{array}{l}{\log }_8\left(x^2-4x+3\right)\le 1\\ \iff x^2-4x+3\le 8\text{ and }{x}^2-4x+3>0\\ \iff x^2-4x-5\le 0\text{ and }{x}^2-4x+3>0\\ \iff (x-5)(x+1)\le 0\text{ and }(x-1)(x-3)>0\\ \iff -1\le x\le 5\text{ and }x\notin [1,3]\\ \iff -1\le x<1\text{ or }3<x\le 5\end{array}$