Solving $\log_2 x\log_3 2x + \log_3 x\log_2 4x > 0$ I have following inequality:
$$\log_2 x\log_3 2x + \log_3 x\log_2 4x > 0$$
which I have reduced to:
$$\frac{\log x}{\log 2}\frac{2x}{\log 3} + \frac{\log x}{\log 3}\frac{4x}{\log 2} > 0$$
$$6x \log x > 0$$
For this to be defined, $x > 0$; also, if $x > 1$, then both $6x$ and $\log x$ are positive.
But I do not know how to solve it completely.
 A: Should be: $$\frac{\log_2 x \log_2 2x}{\log_2 3} + \frac{\log_2 x \log_2 4x}{\log_2 3} > 0$$
$$\log_2 x (1+\log_2 x) + \log_2 x (2+\log_2 x) > 0$$
$$\log_2 x \cdot (3+2\log_2 x) > 0$$
now you have to consider two cases: \begin{cases}
\log_2 x >0 \\
3+2\log_2 x >0
\end{cases} and
\begin{cases}
\log_2 x <0 \\
3+2\log_2 x <0
\end{cases}
Can you finish?
A: Hint
Be carefull, your idea should be translated to
$$\frac{\log x}{\log 2}\frac{\log 2x}{\log 3} + \frac{\log x}{\log 3}\frac{\log 4x}{\log 2} > 0$$
Next step would be
$$\frac{\log x}{\log 2}\frac{(\log x +\log 2)}{\log 3} + \frac{\log x}{\log 3}\frac{(\log x+\log 4)}{\log 2} > 0$$
remember that $\log 2>0$ and $\log 3 >0$. Can you finish?
A: $\begin{cases}
\log_2x>0\\
3+2\log_2x>0
\end{cases}\quad\lor$
$\quad\begin{cases}
\log_2x<0\\
3+2\log_2x<0
\end{cases}$
$\begin{cases}
x>1\\
\log_2x>-\dfrac32
\end{cases}\quad\lor$
$\quad\begin{cases}
0<x<1\\
\log_2x<-\dfrac32
\end{cases}$
$\begin{cases}
x>1\\
x>2^{-\frac32}
\end{cases}\quad\lor$
$\quad\begin{cases}
0<x<1\\
0<x<2^{-\frac32}
\end{cases}$
$\begin{cases}
x>1\\
x>\dfrac1{\sqrt8}
\end{cases}\quad\lor$
$\quad\begin{cases}
0<x<1\\
0<x<\dfrac1{\sqrt8}
\end{cases}$
$\begin{cases}
x>1\\
x>\dfrac{\sqrt2}4
\end{cases}\quad\lor$
$\quad\begin{cases}
0<x<1\\
0<x<\dfrac{\sqrt2}4
\end{cases}$
$x>1\quad\lor\quad0<x<\dfrac{\sqrt2}4\;.$
Have you managed to finish now?
