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? 

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.
