Stuck on solving $\log x - 1 > \frac{2}{\log x}$

Question

I'm stuck in this $$\log x - 1 > \frac{2}{\log x}$$

The book solution is

$$1/10<x<1$$ $$x>100$$

What I've tried so far:

1^st attempt

$$x>0$$

\begin{eqnarray} \log^2x - \log x > 2 \\ \log^{2}x - \log x - 2> 0 \end{eqnarray}

So assuming

$$T = \log x$$

$$T^2 - T - 2 > 0$$

\begin{eqnarray} x1 = 2\\ x2 = -1 \end{eqnarray}

Substituting

$$\log x=2$$

$$4 - 2 - 2 > 0 $$

Substituting

$$\log x=-1$$

$$1 + 1 - 2 > 0 $$

2^nd attempt

\begin{eqnarray} \log x - \log 10 > \frac{\log 100}{\log x}\\ \\ \log \frac{x}{10} > \frac{\log 100}{\log x}\\ \\ \frac{x}{10} = \frac{100}{x}\\ \\ x^2=1000 \\ \end{eqnarray}

As you can see no one of attempts gave me a good results.

Answer 1

The following statements are equivalent:

• $\log x-1>\frac{2}{\log x}$

• $\frac{\left(\log x\right)^{2}-\log x-2}{\log x}>0$

• $\left[\left(\log x\right)^{2}-\log x-2\right]\log x>0$

• $\left(\log x-2\right)\left(\log x+1\right)\log x>0$

• $-1<\log x<0\vee\log x>2$

• $10^{-1}<x<10^{0}\vee x>10^{2}$

Answer 2

Case $\#1:$ If $\log x>0$

$$\log x-1>\dfrac2{\log x}\iff0<(\log x)^2-\log x-2=(\log x-2)(\log x+1)$$

$\implies$ either $\log x<-1$ which is impossible

Else $\log x>2\iff x>10^2$

Case $\#2:$ If $\log x<0,$

$$\log x-1>\dfrac2{\log x}\iff0>(\log x)^2-\log x-2=(\log x-2)(\log x+1)$$

$\implies-1<\log x<2$

But we already have $\log x<0$

$\implies-1<\log x<0\iff10^{-1}<x<10^0$

Case $\#3:$ What if $\log x=0$