Question

$|x-1{|}^A=(x-1{)}^7$, where $A=lo{g}_3{x}^2-2lo{g}_{x}9.$

Answer 1

Domain of $x=(0,\infty )-\left\{1\right\}$

${|x-1|}^{{\log }_3{x}^2-2{\log }_{x}9}={(x-1)}^7$

$\Rightarrow {(x-1)}^{2({\log }_{3}x-{\log }_{x}9)}={(x-1)}^7$

$\Rightarrow 2({\log }_{3}x-{\log }_{x}9)=7$

$\Rightarrow 2({\log }_{3}x-2{\log }_{x}3)=7$

$\Rightarrow 2({\log }_{3}x-\frac{2}{{\log }_{3}x})=7$ sicne ${\log }_{a}x=\frac{1}{{\log }_{x}a}$

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

$\Rightarrow 2(t-\frac{2}{t})=7$

$\Rightarrow 2\left(\frac{t^2-2}{t}\right)=7$

$\Rightarrow 2t^2-7t-4=0$

$\Rightarrow (2t+1)(t-4)=0$

$\Rightarrow t=\frac{-1}{2}$,$t=4$

$\Rightarrow {\log }_{3}x=\frac{-1}{2}$,${\log }_{3}x=4$

$\Rightarrow x=\frac{1}{\sqrt{3}},81$