The question is


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?

  • $\begingroup$ Do you mean $\log_4 x + 3$ or $\log_4 (x + 3)$? $\endgroup$ – N. F. Taussig Jul 23 '16 at 13:10
  • $\begingroup$ @N.F.Taussig did correction. $\endgroup$ – 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: $(-3,-2)$.

  • 1
    $\begingroup$ I think cross multiplication is not good in solving inequality . $\endgroup$ – Aakash Kumar Jul 23 '16 at 14:55
  • $\begingroup$ The first step is not legitimate. $\endgroup$ – user376343 Nov 14 '18 at 21:37
  • $\begingroup$ @ervx upload $-2.5$ into the given inequality... $\endgroup$ – 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 $(-1,\infty).$


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