Problem 813

Solution. Since the base of the logarithm is less than one, the given inequality is equivalent to the inequality

$$\log _{6} \frac{x^{2}+x}{x+4}>1$$

which, considering that the base of the logarithm is greater than one, is equivalent to the inequality

$$\frac{x^{2}+x}{x+4}>6$$

i.e., the inequality

$$\frac{x^{2}+x}{x+4}-6>0 \Leftrightarrow \frac{x^{2}-5 x-24}{x+4}>0$$

Since $x^{2}-5 x-24=(x+3)(x-8)$, solving the inequality

$$\frac{(x+3)(x-8)}{x+4}>0$$

by the method of intervals, we find that the set of all its solutions, and thus of the original inequality, is the intervals $-4<x<-3$ and $8<x<+\infty$.

Solving an inequality of the form

$$f\left(\log _{a} x\right) \geqslant 0$$

where $f$ is some function, by the substitution $t=\log _{a} x$ reduces to solving the inequality $f(t) \geqslant 0$ and then solving the corresponding simple logarithmic inequalities.