Question

Solve for ${}^{\prime }{x}^{\prime }$
$2{\log }_3(x-2)+{\log }_3{(x-4)}^2=0$

• A. $3\pm \sqrt{2}$ and $7$
• B. $5\pm \sqrt{2}$ and $3$
• C. $3\pm \sqrt{2}$ and $3$
• D. None of these

Answer 1

Answer: (D)

None of these

$2lo{g}_3(x-2)+lo{g}_3(x-4{)}^2=0$

$\therefore lo{g}_3(x-2{)}^2+lo{g}_3(x-4{)}^2=0$

$\therefore (x-2{)}^2(x-4{)}^2=1$

$\therefore x^2-6x+8=1or{x}^2-6x+8=-1$

$\therefore x^2-6x+7=0or{x}^2-6x+9=0$

$\therefore x=\frac{6\pm \sqrt{36-4\left(7\right)}}{2}or(x-3{)}^2=0$

$\therefore x=\frac{6\pm 2\sqrt{2}}{2}orx=3$

$\therefore x=3\pm \sqrt{2},3$

But for $x=3-\sqrt{2}$ , $lo{g}_3(x-2)$ is not defined

so x=$3+\sqrt{2},3$