how to proceed next in this logarithmic inequality?

Question

The question is

$$\frac{1}{\log_4{\left(\frac{x+1}{x+2}\right)}}<\frac{1}{\log_4{(x+3)}}$$

I did the first step for defining the arguments of both sides and got $x\in(-3,-2)\cup (-1,\infty)$

next I did reciprocal both sides and then what to do?

Answer 1

The inequality holds iff

$$ \log_{4}\left(\frac{x+1}{x+2}\right)>\log_{4}(x+3). $$ Applying the strictly increasing function $4^{x}$ to both sides, we see that the above holds iff

$$ \frac{x+1}{x+2}>x+3. $$

Now note that id $x\in (-1,\infty)$, $\frac{x+1}{x+2}<1$, while $x+3>2$, so the inequality fails. Thus, we only have to consider $x\in (-3,-2)$. In this case, $x+2<0$, hence, the above inequality is equivalent to

$$ x+1<(x+3)(x+2)=x^{2}+5x+6\iff 0<x^{2}+4x+5. $$

Using the quadratic formula, we find that $x^{2}+4x+5$ has no real roots. Hence, the latter inequality above always holds, and, thus, the answer is: $(-3,-2)$.