Answer
Verified
402.6k+ views
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.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE