7.078. $8^{\frac{2}{x}}-2^{\frac{3 x+3}{x}}+12=0$.

7.078. $8^{\frac{2}{x}}-2^{\frac{3 x+3}{x}}+12=0$.
Solution.

Domain of definition: $x \neq 0$.

Rewrite the equation as

$$
2^{\frac{6}{x}}-2^{3+\frac{3}{x}}+12=0,\left(2^{\frac{3}{x}}\right)^{2}-8 \cdot 2^{\frac{3}{x}}+12=0
$$

Solving this equation as a quadratic in terms of $2^{\frac{3}{x}}$, we get

$$
\begin{aligned}
& \left(2^{\frac{3}{x}}\right)_{1}=2, \text { hence }\left(\frac{3}{x}\right)_{1}=1, x_{1}=3, \text { or }\left(2^{\frac{3}{x}}\right)_{2}=6 \text {, hence } \\
& \left(\log _{2} 2^{\frac{3}{x}}\right)_{2}=\log _{2} 6,\left(\frac{3}{x}\right)_{2}=\log _{2} 6, x_{2}=\frac{3}{\log _{2} 6}=3 \log _{6} 2=\log _{6} 8
\end{aligned}
$$

Answer: $3 ; \log _{6} 8$.