The Power of Logarithms

$(2x)^{\log x} = 8^{\log 16}\\ \log (2x)^{\log x} = \log 8^{\log 16}\\ (\log x)(\log 2 + \log x) = (\log 16)(\log 8)\\ (\log x)^2 + (\log 2)(\log x) - (\log 2^4)(\log 2^3) = 0\\ (\log x)^2 + (\log 2)(\log x) - (4 \log 2)(3 \log 2) = 0\\ (\log x + 4 \log 2)(\log x - 3 \log 2) = 0\\ \log x = -4 \log 2,\;3 \log 2\\ \log x = \log 2^{-4},\;\log 2^{3}\\ x= 2^{-4},\;2^3\\ x=\dfrac{1}{16},\;8$

Therefore, the product of the solutions $=\dfrac{1}{16} \times 8 = \dfrac{1}{2} = \boxed{0.5}$

$\begin{aligned}(2x)^{\log x}=&8^{\log16}\\2^{\log x}\cdot x^{\log x}=&(2^3)^{4\log2}\\x^{\log x}=&\dfrac{2^{12\log2}}{2^{\log x}}\\\log x^{\log x}=&\log2^{12\log2-\log x}\\(\log x)(\log x)=&(12\log2-\log x)(\log2)\\\log^2x+\log x\cdot\log2-12\log^22=&0\\(\log x+4\log2)(\log x-3\log2)=&0\\\log x=&\log2^{-4},\log2^3\\x=&\dfrac1{16},8\end{aligned}$

Thus, the product of all solutions is $\dfrac1{16}\times8=\dfrac12=\boxed{0.5}$