Question

If ${\log }_3(x^3-x^2-x+1)-{\log }_3(x-1)-{\log }_3(x+1)=2$, then $x=$

• A. $4$
• B. $10$
• C. $9$
• D. $6$

Answer 1

The correct option is B $10$

${\log }_3(x^3-x^2-x+1)-{\log }_3(x-1)-{\log }_3(x+1)=2$
$\Rightarrow x^3-x^2-x+1>0,$ $x-1>0$ and $x+1>0$
$\Rightarrow (x-1{)}^2(x+1)>0,$ $x>1$ and $x>-1$
$\Rightarrow x\in (1,\infty )$

Now,
${\log }_3\left(\frac{x^3-x^2-x+1}{(x-1)(x+1)}\right)=2$
$\Rightarrow \frac{(x^3-x^2-x+1)}{x^2-1}=9\Rightarrow \frac{(x-1)(x^2-1)}{(x^2-1)}=9\Rightarrow x=10$