Finding X from Exponential Equations $$2^x \cdot 4^{1-x}= 8^{-x}$$

I wrote all the base numbers as a power of 2 but I'm not sure what to do after.
$$2^x \cdot 4^{1-x}= 8^{-x}$$
$$\implies 2^x\cdot (2^2)^{1-x}=(2^3)^{-x}$$
$$\implies2^x\cdot 2^{2-2x}=2^{-3x}$$
$$\implies 2^{x+2-2x}=2^{-3x}$$
$$\implies 2^{2-x}=2^{-3x}$$
comparing both side we have,
$$2-x=-3x$$
$$\implies -2x=2$$
$$\implies x=-1$$
If you are referring to $2^x * 4^{1-x} = 8^{-x}$ then here are the steps

$2^x * 4^{1-x} = 8^{-x}$

$2^x * 4^1 4^{-x} = 8^{-x}$

$2^x * 4\frac{1}{4^{x}} = \frac{1}{8^{x}}$

$4=\frac{4^x}{8^x 2^x}$

$4={\frac{1}{4}}^x$

$log_{\frac{1}{4}}(4)=x$

$x=-1$

Hope this helps!