1. Stating the problem: Solve the equation $\log_{2x}(4) = \log_x(2)$.\n\n2. Recall the change of base formula for logarithms: $\log_a b = \frac{\log b}{\log a}$. We will use natural logarithm $\ln$ for convenience.\n\n3. Rewrite both sides using the change of base formula:\n$$\log_{2x}(4) = \frac{\ln 4}{\ln (2x)} \quad \text{and} \quad \log_x(2) = \frac{\ln 2}{\ln x}.$$\n\n4. Set the equation:\n$$\frac{\ln 4}{\ln (2x)} = \frac{\ln 2}{\ln x}.$$\n\n5. Cross multiply to solve for $x$:\n$$\ln 4 \cdot \ln x = \ln 2 \cdot \ln (2x).$$\n\n6. Use the property $\ln (2x) = \ln 2 + \ln x$:\n$$\ln 4 \cdot \ln x = \ln 2 \cdot (\ln 2 + \ln x) = \ln 2 \cdot \ln 2 + \ln 2 \cdot \ln x.$$\n\n7. Substitute $\ln 4 = 2 \ln 2$ (since $4 = 2^2$):\n$$2 \ln 2 \cdot \ln x = (\ln 2)^2 + \ln 2 \cdot \ln x.$$\n\n8. Rearrange terms to isolate $\ln x$:\n$$2 \ln 2 \cdot \ln x - \ln 2 \cdot \ln x = (\ln 2)^2.$$\n\n9. Factor $\ln x$ on the left side:\n$$\ln x (2 \ln 2 - \ln 2) = (\ln 2)^2.$$\n\n10. Simplify inside the parentheses:\n$$\ln x (\ln 2) = (\ln 2)^2.$$\n\n11. Divide both sides by $\ln 2$ (not zero):\n$$\cancel{\ln x} \cancel{\ln 2} = \frac{(\ln 2)^2}{\cancel{\ln 2}} \Rightarrow \ln x = \ln 2.$$\n\n12. Exponentiate both sides to solve for $x$:\n$$x = e^{\ln 2} = 2.$$\n\n13. Check the domain: $x > 0$ and $2x > 0$ are true for $x=2$. Also, bases of logarithms cannot be 1, and $2x = 4 \neq 1$, $x=2 \neq 1$, so solution is valid.\n\nFinal answer: $$\boxed{2}.$$