Question

If $A=\{x:x^{{\log }_{\sqrt{x}}2x}=4\}$, then $n\left(A\right)=$

• A. $0$
• B. $1$
• C. $2$
• D. infinitely many

Answer 1

The correct option is A $0$

For $A$ to be defined,
$2x>0$ and $\sqrt{x}>0,\sqrt{x}\ne 1$
$\Rightarrow x>0$ and $x\ne 1$
$\therefore x\in (0,\infty )-\left\{1\right\}\cdots \left(1\right)$

Now, $x^{{\log }_{\sqrt{x}}2x}=4$
$\Rightarrow x^{{\log }_x(2x{)}^2}=4$
$\Rightarrow 4x^2=4[\because a^{{\log }_{a}m}=m]$
$\Rightarrow x=\pm 1$
But neither $x=1$ nor $x=-1$ satisfies $\left(1\right)$.