Question

The solution set of the equation ${{\log }_{\dfrac{5}{\left( {{x}^{2}}+6 \right)}}}\left( 13+4{{x}^{2}}-4x \right)={{\log }_{7+\sqrt{x}}}\left( -2x-{{x}^{2}} \right)$ is,
A. 9
B. 16
C. 81
D. None of these

Answer 1

Hint: We will be using the basic concept of logarithms. Also, we will be using the concept of domain of a function to solve the problem. We will be using the fact that for a function ${{\log }_{a}}b$ the valid values of and b are $a>0\ and\ a\ne 1,b>0$.

Complete step-by-step answer:
Now we have to solve the equation, ${{\log }_{\dfrac{5}{\left( {{x}^{2}}+6 \right)}}}\left( 13+4{{x}^{2}}-4x \right)={{\log }_{7+\sqrt{x}}}\left( -2x-{{x}^{2}} \right)$.
To solve the question we will not be directly finding the value of x which satisfies it instead. We will be finding the range of values of x in its domain to see if the solution exists or not in the options given to us.
So, we know that in ${{\log }_{a}}b$
$a>0\ and\ a\ne 1,b>0$
So, using this for both L.H.S we can find the domain of the equation.
So, taking L.H.S we have,
${{\log }_{\dfrac{5}{\left( {{x}^{2}}+6 \right)}}}\left( 13+4{{x}^{2}}-4x \right)$
So by using the properties of logarithm as we discussed above so, we have to find the values of x for which,
$4{{x}^{2}}-14x-13>0$
and
$\begin{align}
  & \dfrac{5}{{{x}^{2}}+6}>0\forall x\in R\ \And \ \dfrac{5}{{{x}^{2}}+6}\ne 1\forall x\in R \\
 & 13+4{{x}^{2}}-4x=4{{x}^{2}}-4x+13 \\
 & =4{{x}^{2}}-4x+13 \\
\end{align}$
Now, we will find D (discriminant) of the quadratic equation,
$\begin{align}
  & D={{\left( -4 \right)}^{2}}-4\left( 4 \right)\left( 13 \right) \\
 & =16-16\times 13 \\
 & =16\left( 1-13 \right) \\
 & =-12\times 16 \\
 & \Rightarrow D<0 \\
\end{align}$
Since, D < 0 and the coefficient of ${{x}^{2}}$ i.e. 4 > 0 therefore we know that the curve will be above x-axis for all values of x therefore,
$4{{x}^{2}}-4x+13>0\ \forall x\in R$
Since both the inequalities are satisfied for all values of x in left hand side so we have
$x\in R$………..(1) for left hand side
Now, taking R.H.S we again apply the same rule of domain of logarithmic function so we have
Now we have,
$\begin{align}
  & {{\log }_{7+\sqrt{x}}}\left( -2x-{{x}^{2}} \right) \\
 & 7+\sqrt{x}>0\ \forall x\in R \\
 & 7\sqrt{x}\ne 1 \\
 & \sqrt{x}\ne 1-\sqrt{7} \\
 & x\ne {{\left( 1-\sqrt{7} \right)}^{2}} \\
 & \Rightarrow x\in R-\left\{ {{\left( 1-\sqrt{7} \right)}^{2}} \right\}...........\left( 2 \right) \\
 & Also, \\
 & -2x-{{x}^{2}}>0 \\
 & -x\left( 2+x \right)>0 \\
 & x\left( 2+x \right)<0............\left( 3 \right) \\
\end{align}$
Now we have x not equal to ${{\left( 1-\sqrt{7} \right)}^{2}}$by solving the base of the logarithm for domain so we have x belongs to all values of R except ${{\left( 1-\sqrt{7} \right)}^{2}}$in the domain of the base of the logarithm. Now for $x\left( 2+x \right)<0$.We have Signs of inequality change because of negative signs.
Now on the number line we will draw the nature of $x\left( 2+x \right)$ in terms of + and – to solve (3).

Now, by putting suitable value of x in (3)
As $x<-2\ \ \ \ \ (3)$ is positive
As $-2\ \le x<0\ \ \ \ (3)$ is negative
As $x>0\ \ \ \ \ (3)$ is positive
So, we have $x\left( 2+x \right)<0\ \forall x\in \left( -2,0 \right)..........\left( 4 \right)$
Now if we combine both the inequalities (2) and (4) we have that the domain of the function is only the intersection of both (2) and (4) so we have x belongs to (-2, 0) for R.H.S.
Now in L.H.S we have the domain as $x\in R$but for R.H.S we have the domain as (-2, 0) so, we see that the domain of the given question is (-2, 0) as we have to take the intersections of the domains of both the logarithmic functions so the valid domain by doing this is (-2, 0) as only this much part is common in all the domains. Now we can see that option (a), (b), (c) lie outside its domain. Hence they can’t be true.
Therefore the answer is option (D).

Note:These types of questions are best solved through option elimination. Hence, it is advised to check the domain in such questions. Also it should be noted that we have used the domain of logarithmic functions to form inequalities and solve them to find the valid domain of the function which helped us to shorten the solution.