Reverse Logarithmic Inequality Given
$$\frac{1}{\log_{4}\left(\frac{x+1}{x+2}\right)}<\frac{1}{\log_{4}(x+3)}.$$
Then what is the range of values of $x$ for which this inequality is satisfied. 
My Try On simplification, I got $x$ from $(-\infty,-2)$. That however is not the correct answer. Can someone tell me each and every step and every individual domains and inequalities I need to consider in this? 
 A: The arguments of the logarithms should be positive. Hence $x+3>0$ and $\frac{x+1}{x+2}>0$, that is $x\in (-3,-2)\cup (-1,+\infty).$
If $x\in (-3,-2)$ then $0<x+3<1$
and $\frac{x+1}{x+2}>1$. Therefore
$$\log_4(x+3)<0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)>0$$ 
and the inequality does not hold.
If $x\in (-1,+\infty)$ then then $x+3>1$
and $0<\frac{x+1}{x+2}<1$. Therefore
$$\log_4(x+3)>0\quad\mbox{and}\quad\log_{4}\left(\frac{x+1}{x+2}\right)<0$$ 
and the inequality holds.
So the inequality holds iff $x>-1$. Note that if we replace the base $4$ with another number greater than $1$ then the result is the same.
A: HINT: you must do some case work:
$$\log_4(x+3)>0$$ it is $$x+3>1$$ or $$x>-2$$
$$\log_4\left(\frac{x+1}{x+2}\right)>\log_41$$ this is equivalent to $$\frac{x+1}{x+2}>1$$ which is not true.
can you solve the other cases?
if $$\log_4(x+3)<\log_4 1$$ we get $$x<-2$$
if $$\log_4\left(\frac{x+1}{x+2}\right)>\log_41$$ we get
$$\frac{x+1}{x+2}>1$$ and we get $$x+1<x+2$$ which is true.
and you have to solve $$x+3>\frac{x+1}{x+2}$$ this is equivalent to
$$(x+3)(x+2)<x+1$$ this gives $$x^2+4x+5<0$$ or $$(x+1)^2+1<0$$ which is impossible.
A: We can use the intervals method.
The domain gives $x>-1$ or $-3<x<-2$.
We need to solve that
$$\frac{1}{\log_4(x+3)}-\frac{1}{\log_4\frac{x+1}{x+2}}>0$$ or
$$\frac{\log_4\frac{x+1}{(x+2)(x+3)}}{\log_4(x+3)\log_4\frac{x+1}{x+2}}>0.$$
Now, since $\frac{x+1}{(x+2)(x+3)}\neq1$, it's enough to check two intervals only, which gives the answer:
$$(-1,+\infty).$$
