Solving $\log_x(x^2+4x)>-1$

Question

I'm stuck looking for a solution for this. Any hint? $$ \log_x(x^2+4x) > -1 $$

It looks like $$ x^2+4x > 1/x $$ which I cannot solve.

Answer 1

Hints:

$$\log_x(x^2+4x)>-1=\log_xx^{-1}\implies \begin{cases}x^2+4x>\frac1x&,\;\;\text{if}\;\;x>1\\{}\\x^2+4x<\frac1x&,\;\;\text{if}\;\;0<x<1\end{cases}$$