Question

Solve the following inequation and find the values of ‘x’ which satisfy it: ${{\log }_{{}^{1}/{}_{5}}}\left( 2{{x}^{2}}+5x+1 \right)<0$

Answer 1

Hint: To solve the question given above, first we will equate the term $\left( 2{{x}^{2}}+5x+1 \right)$ to ‘t’ then we will find the range of values of ‘t’ which we can use inside the logarithm i.e. we will find domain of ‘t’ . Then we will use the log remaining identity to remove the logarithm from the left hand side of the inequality. Then after removing the logarithm, we will change the sign of inequality because the base of the logarithm is less than 1 and then we will solve the inequality and find its intersection with the domain.

Complete step-by-step answer:
First, we will equate the term $\left( 2{{x}^{2}}+5x+1 \right)$ to ‘t’. Thus, we get:
${{\log }_{\dfrac{1}{5}}}t<0$
Now, we know that, we cannot put any value of ‘t’ in the above equation. In the above inequation, ‘t’ can have only positive values. Thus, we have:
$t>0$
$\Rightarrow 2{{x}^{2}}+5x+1=0.......\left( 1 \right)$
Now, we have to solve the above inequation and to find the values of ‘x’ which will satisfy the above inequation. In simple words, we have to find the domain of ‘x’. As we can see that we cannot factorise $\left( 2{{x}^{2}}+5x+1 \right)$ , we will assume that its roots are $\alpha$ and $\beta$. Thus, we will get following equation:
$\left( x-\alpha \right)\left( x-\beta \right)>0......\left( 2 \right)$
The solution of the above inequation is $\left( -\infty ,\alpha \right)\cup \left( \beta ,\infty \right)$ where $\alpha <\beta$. Now we will find the values of $\alpha$ and $\beta$. As $\alpha$ and $\beta$ are the roots of $2{{x}^{2}}+5x+1=0,$ we can calculate them with the help of quadratic formula. The quadratic formula for the calculation of roots of $a{{x}^{2}}+bx+c=0$ is :
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Thus, we have:
$x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 2 \right)\left( 1 \right)}}{2\left( 2 \right)}$
$\Rightarrow x=\dfrac{-5\pm \sqrt{25-8}}{4}$
$\Rightarrow x=\dfrac{-5\pm \sqrt{17}}{4}$
So, we can say that, $\alpha =\dfrac{-5-\sqrt{17}}{4}$ and $\beta =\dfrac{-5+\sqrt{17}}{4}$. So the solution of the equation (2) will become:
$x\in \left( -\infty ,\dfrac{-5-\sqrt{17}}{4} \right)\cup \left( \dfrac{-5+\sqrt{17}}{4},\infty \right)...........\left( 3 \right)$
Now, as we have obtained the domain of ‘x’, we will proceed by solving the inequality given below:
${{\log }_{\left( \dfrac{1}{5} \right)}}\left( 2{{x}^{2}}+5x+1 \right)<0$
Now, we will use the logarithm removing identity which is as shown below:
${{\log }_{a}}b=c\Rightarrow b={{a}^{c}}$
Thus, we will get:
$\left( 2{{x}^{2}}+5x+1 \right)>{{\left( \dfrac{1}{5} \right)}^{0}}$
Here, we have changed the sign of inequality because the value of $\left( \dfrac{1}{5} \right)$ is less than ‘c’ so whenever the value of base is less than 1 and we apply the logarithm remaining identity, we have to change the sign of inequality.
Thus, we will get:
$2{{x}^{2}}+5x+1>1$
$\Rightarrow 2{{x}^{2}}+5x>0$
$\Rightarrow x\left( 2x+5 \right)>0$
$\Rightarrow x\in \left( -\infty ,\dfrac{-5}{2} \right)\cup \left( 0,\infty \right)....\left( 4 \right)$
Now the final solution of the inequation will be obtained by doing intersection of (3) and (4) i.e. we will take only those values of ‘x’ which is common to both (3) and (4).
Before doing this, we will calculate the values of $\alpha$ and $\beta$ in decimal form. So, $\alpha =\dfrac{-5-\sqrt{17}}{4}=\dfrac{-5-4.12}{4}=\dfrac{-9.12}{4}$
$\Rightarrow \alpha =-2.28$
$\beta =\dfrac{-5+\sqrt{17}}{4}=\dfrac{-5+4.12}{4}=\dfrac{-0.88}{4}=-0.22$
So the domain becomes: $x\in \left( -\infty ,-2.28 \right)\cup \left( -0.22,\infty \right)$. Now we will do an intersection. So after intersection, we will get:
$x\in \left( -\infty ,\dfrac{-5}{2} \right)\cup \left( 0,\infty \right)$

Note: In this case, we can see that even after the intersection with the domain, the final answer remains the same as the answer of inequality. However, this is not always the case. It may happen that we do not get the same answer. It may also happen that we do not get any value of ‘x’ that satisfies the particular inequation.