Question

The set of all x satisfying the equation $x^{{\log }_3{x}^2+{\left({\log }_{3}x\right)}^2-10}=\frac{1}{x^2}$

• A. $1,9$
• B. $1,9,1/81$
• C. $1,4,1/81$
• D. $9,1/81$

Answer 1

The correct option is B $1,9,1/81$

$x^{{\log }_3{x}^2+{\left({\log }_{3}x\right)}^2-10}=\frac{1}{x^2}$

Taking log of both the side with base $3$, we have

$({\log }_3{x}^2+{\left({\log }_{3}x\right)}^2-10)\left({\log }_{3}x\right)=-2{\log }_{3}x\cdots \cdots [\because a^{{\log }_{a}b}=b]$.

$({\log }_3{x}^2+{\left({\log }_{3}x\right)}^2-10)\left({\log }_{3}x\right)+2{\log }_{3}x=0$.

$({\log }_3{x}^2+{\left({\log }_{3}x\right)}^2-8)\left({\log }_{3}x\right)=0$

$\Rightarrow {\log }_{3}x=0$ or $\Rightarrow 2{\log }_{3}x+{\left({\log }_{3}x\right)}^2-8=0$.

$\Rightarrow x=1,{\log }_{3}x=\frac{-2\pm \sqrt{4+4\times 1\times 8}}{2}=-1\pm 3$ or ${\log }_{3}x=2,{\log }_{3}x=-4$

So,$x=1,9,\frac{1}{81}$

So, option $B$ is correct.