Explanation
$$f(x)=\log _{ e }{ \{ (x+1)(x+2)\} } $$
For the function to be defined
$$(x+1)(x+2)>0\\ \Rightarrow -2>x>-1\\ \Rightarrow x\in \left( -\infty ,-2 \right) \cup \left( -1,\infty \right) $$
$$9{ x }^{ 2 }-18|x|+5=0\\ 9{ |x| }^{ 2 }-18|x|+5=0\\ \Rightarrow 9|{ x| }^{ 2 }-18|x|+5=0\quad for(x>0)\\ \Rightarrow 9|{ x| }^{ 2 }-3|x|-15|x|+5=0\\ \Rightarrow 3|x|(3|x|-1)-5(3|x|-1)=0\\ \Rightarrow (3|x|-1)(3|x|-5)=0\\ \Rightarrow |x|=\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 3 } \\ \Rightarrow x=\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 3 } ,-\dfrac { 1 }{ 3 } ,-\dfrac { 5 }{ 3 } $$
Clearly $$-\dfrac { 5 }{ 3 } $$ do not belong to the domain of the function $$f(x)$$
So there are $$3$$ solutions belonging to the domain of $$f(x)$$
Option $$C$$ is ccorrect.
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