Question

If ${\log }_{\left(3x-1\right)}\left(x-2\right)={\log }_{\left(9x^2-6x+1\right)}\left(2x^2-10x-2\right)$ , then $x$ equals

• A. $9-\sqrt{15}$
• B. $3+\sqrt{15}$
• C. $2+\sqrt{5}$
• D. $6-\sqrt{5}$

Answer 1

Answer: (B)

We have,
${\log }_{3x-1}\left(x-2\right)$
$={\log }_{\left(9x^2-6x+1\right)}\left(2x^2-10x-2\right)$
$\Rightarrow {\log }_{\left(3x-1\right)}\left(x-2\right)$
$={\log }_{{\left(3x-1\right)}^2}\left(2x^2-10x-2\right)$
$\Rightarrow {\log }_{3x=1}\left(x-2\right)$
$=\frac{1}{2}{\log }_{3x-1}\left(2x^2-10x-2\right){\left(x-2\right)}^2$
$\Rightarrow {\left(x-2\right)}^2=2x^2-10x-2$
$\Rightarrow x^2-4x+4=2x^2-10x-2$
$\Rightarrow x^2-6x-6=0$
$\Rightarrow x=3\pm \sqrt{15}$
$\Rightarrow x=3-\sqrt{15},3+\sqrt{15}$
At $x=3-\sqrt{15}$
$\Rightarrow x-2<0$
$\therefore x=3+\sqrt{15}$