Question

Find the values of $x$ satisfying the equation ${|x-1|}^{{\log }_3{x}^2-2{\log }_{x}9}=(x-1{)}^7$

• A. $2,81$
• B. $2,-1$
• C. $4,-\frac{1}{4}$
• D. $4,81$

Answer 1

The correct option is A $2,81$

${|x-1|}^{{\log }_3{x}^2-2{\log }_{x}9}={(x-1)}^7$
For above equation to be defined, $x\ne 1$ and $x>1$
Taking logarithm on both sides, we get

$({\log }_3{x}^2-2{\log }_{x}9)\log |x-1|=7\log (x-1)[\because \log a^m=m\log a]$

$\Rightarrow (2{\log }_{3}x-\frac{4}{{\log }_{3}x})\log |x-1|=7\log (x-1)[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
For $x>1,|x-1|=(x-1)$

$\Rightarrow (2{\log }_{3}x-\frac{4}{{\log }_{3}x})\log (x-1)=7\log (x-1)$

$\Rightarrow \log (x-1)=0$ $\Rightarrow x=2$

or $2{\left({\log }_{3}x\right)}^2-7{\log }_{3}x-8=0$

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

$\Rightarrow x=81,\frac{1}{\sqrt{3}}$
Since, $x>1$
Therefore, $x=2,81$