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

Making the base 5, we have

${\log }_5\left(16-x^2\right)-{\log }_5(2x+5)\le {\log }_{5}5$

$\therefore \frac{16-x^2}{2x+5}\le 5$ or $\left(16-x^2\right)\le 10x+25$

or $x^2+10x+9\ge 0$

or $(x+9)(x+1)\ge 0\therefore x\le -9$ or $x\ge -1$

Now by definition of log, we must have  …(1)

$2x+5>0,\text{ i.e. },x>-5/2$ and $16-x^2>0$or $x^2-16<0$ or $(x+4)(x-4)<0$

$\therefore -4<x<4$ and $x>-5/2$      ... (2)

Hence from (1) and (2), $x\ge -1$ and x<4.

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