how to proceed next in this logarithmic inequality?

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?

• Do you mean $\log_4 x + 3$ or $\log_4 (x + 3)$? – N. F. Taussig Jul 23 '16 at 13:10
• @N.F.Taussig did correction. – danny Jul 23 '16 at 13:13

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: $(-3,-2)$.

• I think cross multiplication is not good in solving inequality . – Aakash Kumar Jul 23 '16 at 14:55
• The first step is not legitimate. – user376343 Nov 14 '18 at 21:37
• @ervx upload $-2.5$ into the given inequality... – user376343 Nov 15 '18 at 8:05

Consider $$x\in(-3,-2)\cup (-1,\infty)$$. Denote this set $$\mathcal{D}.$$

Since we are working with an inequality check the signs of both logarithms.

1. Solve $$\frac{x+1}{x+2}>1 \;\text{and}\;(x+3)>1.$$ We get $$(x<-2)$$ and $$(x>-2),$$ which gives an empty set. The logarithms in the given equation cannot be simultaneously positive.

2. Similarly, solving $$\frac{x+1}{x+2}<1 \;\text{and}\;(x+3)<1$$ leads to $$(x>-2)$$ and $$(x<-2),$$ again an empty set.

3. It remains the possibility when one logarithm is positive and one negative. From $$\frac{1}{\log_4{\left(\frac{x+1}{x+2}\right)}}<\frac{1}{\log_4{(x+3)}}$$ it is clear that $$\log_4{\left(\frac{x+1}{x+2}\right)}<0$$ and $$\log_4{(x+3)}>0.$$ This holds iff $$x>-2$$ and $$x\in \mathcal{D}.$$

The set of solutions is $$(-1,\infty).$$