Solve the logarithmic equation:

$\log_{x}(2(1-x)) = \log_{1-x} (2x)$

$\dfrac{\log (2(1-x))}{\log x} = \dfrac{\log (2x)}{\log (1-x)}$

$(\log 2+\log (1-x))(\log (1-x)) = (\log 2+\log x)(\log x)$

$\log^{2} (1-x) - \log^{2} x + (\log 2)(\log (1-x)-\log x) = 0$

$(\log (1-x)-\log x)(\log (1-x)+\log x+\log 2) = 0$

• If $\log (1-x)-\log x = 0$ , then $1-x = x \implies \boxed{x = 0.5}$ .
• If $\log (1-x)+\log x+\log 2 = 0$ , then $2x^{2}-2x+1 = 0$ . This equation has no real root.

Therefore, the answer is $x = 0.5$ .

$\begin{aligned} \log_x (2(1-x)) - \log_{1-x}(2x) & = 0 \\ \implies \log_{\color{#D61F06}x} (2{\color{#3D99F6}(1-x)}) & = \log_{\color{#3D99F6}1-x}(2{\color{#D61F06}x}) \end{aligned}$

Equating the logarithm bases and the operants, we have $x=1-x$ $\implies x = \dfrac 12 = \boxed{0.5}$ .