The solution set of the equation  ${\log }_{\frac{1}{5}}(2x+5)+{\log }_5(16-x^2)\le 1$.

$\begin{array}{l}{\log }_{\frac{1}{5}}(2x+5)+{\log }_5(16-x^2)\le 1\\ \Rightarrow {\log }_5(16-x^2)+{\log }_{5^{-1}}(2x+5)\le {\log }_{5}5\\ \Rightarrow {\log }_5\left(\frac{16-x^2}{2x+5}\right)\le {\log }_{5}5\\ \Rightarrow \frac{16-x^2}{2x+5}\le 5\Rightarrow 16-x^2\le 5(2x+5)\\ \Rightarrow 16-x^2-10x-25\le 0\\ \Rightarrow x^2+10x+9\ge 0\\ \Rightarrow (x+9)(x+1)\ge 0\\ \therefore x\le -9\left(or\right)x\ge -1\to \left(1\right)\end{array}$

By definition of log,

We must have  $2x+5>0\Rightarrow x\ge \frac{-5}{2}$

and $16-x^2>0$ (or) $x^2-16<0$

$\Rightarrow (x+4)(x-4)<0$

$\Rightarrow -4<x<4$

$\therefore x\ge \frac{-5}{2}$ and $-4<x<4\to \left(2\right)$

From equations (1) and (2), we get

$x\ge -1\text{\hspace{0.17em}and\hspace{0.17em}}x<4$

$\therefore x\in [-1,4)$.