Question

The value of $x$ which satisfy the equation $(x-1{)}^{{\log }_3{x}^2-2{\log }_{x}9}=(x-1{)}^7$ is equal to

• A. $3,17$
• B. $\frac{1}{\sqrt{3}},81$
• C. $13,21$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{1}{\sqrt{3}},81$

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

$\Rightarrow x-1=0or{\log }_3{x}^2-2{\log }_{x}9=7$

$\Rightarrow x=1$

$or{\log }_3{x}^2-2{\log }_x{3}^2=7$

$\Rightarrow \frac{2}{{\log }_{x}3}-4{{\log }_x}^3=7$

$\Rightarrow Let{\log }_{x}3=y$

$\Rightarrow \frac{2}{y}-4y=7$

$\Rightarrow 2-4y^2=7y$

$\Rightarrow 4y^2+7y-2=0$

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

$\Rightarrow 4y\left(y+2\right)-1\left(y+2\right)=0$

$\Rightarrow \left(y+2\right)\left(4y-1\right)=0$

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

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

$\Rightarrow 3=x^{-1}\Rightarrow 3=x^{\frac{1}{4}}$

$\Rightarrow 3=\frac{1}{x^2}\Rightarrow {\left(3\right)}^4=x$

$\Rightarrow x^2=\frac{1}{3}\Rightarrow x=81$

$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$

Since $x$ cannot be $-\frac{1}{\sqrt{3}}$

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