Solve: \[{{\left( \ln x \right)}^{2}}-\left( \ln 2 \right)\left( \ln x \right)<2{{\left( \ln 2 \right)}^{2}}\]

Hint: We will have to convert the above inequality to a quadratic inequality by putting $\ln x=t$ and then solving the inequality. Also, the formula used to solve quadratic equation $a{{x}^{2}}+bx+c$ is given as: -
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step-by-step answer:
As given in question we have to find the values of x or range of values of x that satisfy the given inequality \[{{\left( \ln x \right)}^{2}}-\left( \ln 2 \right)\left( \ln x \right)<2{{\left( \ln 2 \right)}^{2}}\]. This is a complex type of inequality because here we have terms like $\ln x$ and ${{\left( \ln x \right)}^{2}}$. So, first we have to deal with it first. So, simplify it, we will put $\ln x=t$ ……(i)
After putting this, the inequality given in question becomes
${{t}^{2}}-\ln 2\left( t \right)<2{{\left( \ln 2 \right)}^{2}}$
${{t}^{2}}-\ln 2\left( t \right)<2{{\left( \ln 2 \right)}^{2}}$ …………………………………….(ii)
Thus, we have to find those values of t which when will put in the term ${{t}^{2}}-\ln 2\left( t \right)<2{{\left( \ln 2 \right)}^{2}}$, it will become negative. Above inequality is a quadratic inequality in t. Thus, to solve this, first we will have to find the zeros of the term on the left side of the inequality. Thus,
 ${{t}^{2}}-\ln 2\left( t \right)-2{{\left( \ln 2 \right)}^{2}}=0$ ………………………….(iii)
The zeroes of the above equation can be calculated by quadratic formula. Using quadratic formula, we get
$t=\dfrac{\ln 2\pm \sqrt{{{\left( \ln 2 \right)}^{2}}-4\left( -2 \right){{\left( \ln 2 \right)}^{2}}}}{2\left( 1 \right)}$
$\Rightarrow t=\dfrac{\ln 2\pm \sqrt{{{\left( \ln 2 \right)}^{2}}+8{{\left( \ln 2 \right)}^{2}}}}{2}$
$\Rightarrow t=\dfrac{\ln 2\pm \sqrt{9{{\left( \ln 2 \right)}^{2}}}}{2}$
$\Rightarrow t=\dfrac{\ln 2\pm 3\ln 2}{2}$
$\Rightarrow t=-\ln 2$ and $t=2\ln 2$
Thus, we can write equation (i) in from of this: -
$\left( t+\ln 2 \right)\left( t-2\ln 2 \right)<0$
$\Rightarrow t\in \left( -\ln 2,2\ln 2 \right)$
But actually, we have to find the values of x that satisfies the inequality. So, to get the range in terms of x, we will put value of t i.e., $\ln x$ in above,
$\Rightarrow \ln x\in \left( -\ln 2,2\ln 2 \right)$
$\Rightarrow \ln x\in \left( \ln \left( \dfrac{1}{2} \right),\ln {{\left( 2 \right)}^{2}} \right)$
We have written them in this form because $a\ln x=\ln {{x}^{a}}$.
$\Rightarrow \ln x\in \left( \ln \left( \dfrac{1}{2} \right),\ln 4 \right)$
Also, $\ln x$ is an increasing function, so we can write directly as: -
$x\in \left( \dfrac{1}{2},4 \right)$
Hence, this is our required solution.

Note: Here, we have not made the use of the domain of $\ln x$ according to which x should always be positive. When we will do the intersection of the $x\in \left( 0,\infty \right)$ and $x\in \left( \dfrac{1}{2},4 \right)$, the answer would still be same but if we had negative values in the answer, the final answer should be changed accordingly.