8/( √2)^х=4^(3х-2)*√2

Решите задачу:

$\frac{8}{ (\sqrt{2} )^x} =4^{3x-2}* \sqrt{2} \\ \frac{8}{ (\sqrt{2} )^x} =4^{3x}* 4^{-2}* \sqrt{2} \\\frac{8}{ (\sqrt{2} )^x} =64^{x}* \frac{1}{16} * \sqrt{2} \\ 64^x* (\sqrt{2} )^x= \frac{8*16}{ \sqrt{2}} \\ (64 \sqrt{2} )^x= \frac{128}{\sqrt{2}} \\ (64 \sqrt{2} )^x= \frac{128\sqrt{2}}{\sqrt{2}*\sqrt{2}} \\ (64 \sqrt{2} )^x= \frac{128\sqrt{2}}{2} \\ (64 \sqrt{2} )^x= 64\sqrt{2} \\ (64 \sqrt{2} )^x= (64\sqrt{2})^1 \\ x=1$