Step-by-step Solution:

\[ \log_x 2 \log_{2x} 2 = \log_{4x} 2 \] \[ \therefore \log_x 2 = \frac{1}{\log_{2x} x}, \quad \log_{2x} 2 = \frac{1}{\log_{2x} x}, \quad \text{and} \quad \log_{4x} 2 = \frac{1}{\log_{4x} x} \] \[ \Rightarrow \frac{1}{\log_x 2} * \frac{1}{\log_{2x} x} = \frac{1}{\log_{4x} x} \] \[ \Rightarrow \log_x 2 \log_{2x} x = \log_{2} 4x \] \[ \Rightarrow \log_x 2 (\log_2 2 + \log_x 2) = \log_2 4 + \log_x 2 \] \[ \text{Let } \log_x 2 = a \] \[ \Rightarrow a(1 + a) = \log_2 2^2 + a \quad (\because \log_2 2 = 1) \] \[ \Rightarrow a + a^2 = 2 + a \] \[ \Rightarrow a^2 - a - 2 = 0 \] \[ \Rightarrow a = \pm \sqrt{2} \] \[ \Rightarrow \log_x 2 = \pm \sqrt{2} \] \[ \Rightarrow x = 2^{\sqrt{2}}, 2^{-\sqrt{2}} \]