Answer
$x\displaystyle \in\{\frac{1}{2},5000\}$
Work Step by Step
In order for the equation to be defined,
$\left\{\begin{array}{llll}
2x\geq 0 & \Rightarrow & x\geq 0 & \\
\log 2x\geq 0 & \Rightarrow & 2x\geq 1 & \Rightarrow x\geq\frac{1}{2}\\
2x\gt 0 & \Rightarrow & x\gt 0 &
\end{array}\right. \qquad(*)$
Rewrite $\sqrt{2x}$ as $(2x)^{1/2}$, and apply $ \quad\log_{a}M^{p}=p\cdot\log_{a}M$
$\displaystyle \frac{1}{2}\log 2x=\sqrt{\log 2x}\qquad$ ... let $t=\log 2x$
$\displaystyle \frac{1}{2}t=\sqrt{t}\qquad$ ... square both sides
$\displaystyle \frac{1}{4}t^{2}=t$
$t^{2}-4t=0$
$t(t-4)=0$
Solutions for t: $t=0,t=4.$
Bring back x
$\left[\begin{array}{lll}
\log 2x=0 & ... & \log 2x=4\\
2x=10^{0} & & 2x=10^{4}\\
x=1/2 & & x=5000
\end{array}\right]$
Both satisfy (*), so both are valid solutions.
