# 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.

• Shouldn't that be $$\frac{\log x}{\log 2}\frac{\log 2x}{\log 3} + \frac{\log x}{\log 3}\frac{\log 4x}{\log 2} > 0?$$ – saulspatz Apr 5 at 13:03

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?

• Yes I got it myself. – Shiv Apr 5 at 13:20
• @Sil: since $\log_2 3>0$ we can multiply both sides by it and get rid of it – Vasya Apr 5 at 13:37
• @Sil: Oh, yes, thanks! – Vasya Apr 5 at 13:39

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?

• Yes I got it myself. – Shiv Apr 5 at 13:20
• @Shiv: If you are satisfied, please, choose one as an answer. Doing this you avoid your question popping up in the question list. – Arnaldo Apr 5 at 13:55

$$\begin{cases} \log_2x>0\\ 3+2\log_2x>0 \end{cases}\quad\lor$$ $$\quad\begin{cases} \log_2x<0\\ 3+2\log_2x<0 \end{cases}$$

$$\begin{cases} x>1\\ \log_2x>-\dfrac32 \end{cases}\quad\lor$$ $$\quad\begin{cases} 0

$$\begin{cases} x>1\\ x>2^{-\frac32} \end{cases}\quad\lor$$ $$\quad\begin{cases} 0

$$\begin{cases} x>1\\ x>\dfrac1{\sqrt8} \end{cases}\quad\lor$$ $$\quad\begin{cases} 0

$$\begin{cases} x>1\\ x>\dfrac{\sqrt2}4 \end{cases}\quad\lor$$ $$\quad\begin{cases} 0

$$x>1\quad\lor\quad0

Have you managed to finish now?

• Yes, I have finished it. – Shiv Apr 6 at 14:01