7.229. $27 \cdot 2^{-3 x}+9 \cdot 2^{x}-2^{3 x}-27 \cdot 2^{-x}=8$.
Solution.

Transform the equation:

$$
\begin{aligned}
& 27+9 \cdot 2^{4 x}-2^{6 x}-27 \cdot 2^{2 x}=8 \cdot 2^{3 x} \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-9 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{2 x}-27=0 \Leftrightarrow \\
& \Leftrightarrow 2^{6 x}-2^{4 x}-8 \cdot 2^{4 x}+8 \cdot 2^{3 x}+27 \cdot 2^{x}-27=0 \Leftrightarrow
\end{aligned}
$$

$$
\begin{aligned}
& \Leftrightarrow 2^{4 x}\left(2^{2 x}-1\right)-8 \cdot 2^{3 x}\left(2^{x}-1\right)+27\left(2^{x}-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow 2^{4 x}\left(2^{x}-1\right)\left(2^{x}+1\right)-8 \cdot 2^{3 x}\left(2^{x}-1\right)+27\left(2^{x}-1\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(2^{x}-1\right)\left(2^{5 x}+2^{4 x}-8 \cdot 2^{3 x}+27\right)=0
\end{aligned}
$$

from which $2^{x}=1, x_{1}=0$. The equation $2^{5 x}+2^{4 x}-8 \cdot 2^{3 x}+27=0$ has no solutions.

Answer: 0.