Question

Find the value of $x$, if $x^{{\log }_3\left(x^2\right)+\left[{\log }_3\right(x){]}^2-10}=\frac{1}{x^2}$

Answer 1

Clearly, $x>0$
$x^{{\log }_3\left(x^2\right)+\left[{\log }_3\right(x){]}^2-10}=\frac{1}{x^2}$
Taking logarithm with base $x$ , we get

${\log }_x\left[x^{{\log }_3\left(x^2\right)+\left[{\log }_3\right(x){]}^2-10}\right]={\log }_x{x}^{-2}$

$\left[{\log }_3\right(x^2)+[{\log }_3\left(x\right){]}^2-10]{\log }_{x}x=-2{\log }_{x}x[\because \log x^m=m\log x]$

$\Rightarrow {\log }_3\left(x^2\right)+\left[{\log }_3\right(x){]}^2-10=-2$ ($\because {\log }_{x}x=1$)

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

Put ${\log }_{3}x=t$

$\Rightarrow t^2+2t-8=0$

$\Rightarrow t^2+4t-2t-8=0$

$\Rightarrow t(t+4)-2(t+4)=0$

$\Rightarrow (t-2)(t+4)=0$

$\Rightarrow t=2,-4$

$\Rightarrow {\log }_{3}x=-4$ and ${\log }_{3}x=2$

$\Rightarrow \frac{\log x}{\log 3}=-4$ and $\frac{\log x}{\log 3}=2$

$\Rightarrow \log x=-4\log 3$ and $\log x=2\log 3$

$\Rightarrow x=3^{-4}$ and $x=3^2$

$\Rightarrow x=\frac{1}{81},9$.