Question

If $x^{{\log }_3{x}^2+({\log }_{3}x{)}^2-10}=\frac{1}{x^2}$, then number of value(s) of $x$ satisfying the equation is/are

Answer 1

$x^{{\log }_3{x}^2+({\log }_{3}x{)}^2-10}=\frac{1}{x^2}$
$\Rightarrow x^{2{\log }_{3}x+({\log }_{3}x{)}^2-8}=1(x\ne 0)$
$\Rightarrow 2{\log }_{3}x+({\log }_{3}x{)}^2-8=0\text{or}x=1$
Now, for $2{\log }_{3}x+({\log }_{3}x{)}^2-8=0$
$\text{Let}{\log }_{3}x=t$
$\Rightarrow t^2+2t-8=0\Rightarrow (t-2)(t+4)=0$
$\Rightarrow t=2,-4$
$\Rightarrow {\log }_{3}x=2$ or ${\log }_{3}x=-4$
$\Rightarrow x=9$ or $x=\frac{1}{81}$
$\therefore x=\frac{1}{81},1,9$