Question

Solution set of the inequality ${{\log }_{3}}\left( x+2 \right)\left( x+4 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+2 \right)<\dfrac{1}{2}{{\log }_{\sqrt{3}}}7$ is:
(a) $\left( -2,-1 \right)$
(b) $\left( -2,3 \right)$
(c) $\left( -1,3 \right)$
(d) $\left( 3,\infty \right)$

Answer 1

Hint: In this type of question we have to use the concept of logarithm. WE have to use different properties of logarithm such as ${{\log }_{\dfrac{1}{b}}}a=-{{\log }_{b}}a$, ${{\log }_{b}}a=\dfrac{\log a}{\log b}$, ${{\log }_{{{a}^{b}}}}c=\dfrac{1}{b}{{\log }_{a}}c$, $\log \left( ab \right)=\log a+\log b$ and $\log \left( \dfrac{a}{b} \right)=\log a-\log b$. Here, we consider the given inequality and by using the above properties of logarithm we simplify it to obtain the required set of solutions.

Complete step by step answer:
Now, we have to find the solution set of the inequality ${{\log }_{3}}\left( x+2 \right)\left( x+4 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+2 \right)<\dfrac{1}{2}{{\log }_{\sqrt{3}}}7$
Let us consider the inequality,
$\Rightarrow {{\log }_{3}}\left( x+2 \right)\left( x+4 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+2 \right)<\dfrac{1}{2}{{\log }_{\sqrt{3}}}7\cdots \cdots \cdots \left( i \right)$
As we know, the logarithm is defined for only positive terms. So that, above equation is valid only when
$\Rightarrow \left( x+2 \right)\left( x+4 \right)>0\text{ and }\left( x+2 \right)>0$
$\Rightarrow x>-2\cdots \cdots \cdots \left( ii \right)$
We know that, ${{\log }_{\dfrac{1}{b}}}a=-{{\log }_{b}}a$ so by using this we can write the above inequality as
$\Rightarrow {{\log }_{3}}\left( x+2 \right)\left( x+4 \right)-{{\log }_{3}}\left( x+2 \right)<\dfrac{1}{2}{{\log }_{\sqrt{3}}}7$
Also , we know that, ${{\log }_{{{a}^{b}}}}c=\dfrac{1}{b}{{\log }_{a}}c$, hence by using this we can write the inequality as
$\Rightarrow {{\log }_{3}}\left( x+2 \right)\left( x+4 \right)-{{\log }_{3}}\left( x+2 \right)<{{\log }_{{{\left( \sqrt{3} \right)}^{2}}}}7$
$\Rightarrow {{\log }_{3}}\left( x+2 \right)\left( x+4 \right)-{{\log }_{3}}\left( x+2 \right)<{{\log }_{3}}7$
Now, by using the property of logarithm i.e. $\log \left( ab \right)=\log a+\log b$ we can write the above inequality as
$\begin{align}
  & \Rightarrow {{\log }_{3}}\left( x+2 \right)+{{\log }_{3}}\left( x+4 \right)-{{\log }_{3}}\left( x+2 \right)<{{\log }_{3}}7 \\
 & \Rightarrow {{\log }_{3}}\left( x+4 \right)<{{\log }_{3}}7 \\
\end{align}$
On comparison and further simplification gives us the above inequality as
$\begin{align}
  & \Rightarrow \left( x+4 \right)<7 \\
 & \Rightarrow x<7-4 \\
 & \Rightarrow x<3\cdots \cdots \cdots \left( iii \right) \\
\end{align}$
From equations $\left( ii \right)$ and $\left( iii \right)$ we get,
$\Rightarrow x>-2\text{ and }x<3$
Thus by using interval we can represent the above expression as
$\Rightarrow x\in \left( -2,3 \right)$
Hence, the solution set of the inequality ${{\log }_{3}}\left( x+2 \right)\left( x+4 \right)+{{\log }_{\dfrac{1}{3}}}\left( x+2 \right)<\dfrac{1}{2}{{\log }_{\sqrt{3}}}7$ is $x\in \left( -2,3 \right)$.

So, the correct answer is “Option b”.

Note: In this type of question students have to use logarithm rules for fast exponent calculation. Students should make use of the appropriate logarithmic formulas wherever needed. Also students have to remember that if the base value of logarithm is not written then the base is $e$.