Question

The number of real values of $x$, that satisfy the equation $x^{{\log }_{\sqrt{x}}2x}=4$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$x^{{\log }_{\sqrt{x}}2x}=4$
$\Rightarrow 2x>0,\sqrt{x}>0$ and $\sqrt{x}\ne 1$
$\Rightarrow x\in (0,\infty )-\left\{1\right\}...\left(1\right)$

Now, $x^{{\log }_x(2x{)}^2}=4$
$\Rightarrow (2x{)}^2=4(\because a^{{\log }_{a}b}=b)$
$\Rightarrow x=\pm 1$
From eqn $\left(1\right)$, neither $1$ nor $-1$ satisfies the equation.