7.224. $\frac{8^{x}+2^{x}}{4^{x}-2}=5$
Solution.

Domain of definition: $x \neq \frac{1}{2}$.

Rewrite the equation as $2^{3 x}-5 \cdot 2^{2 x}+2^{x}+10=0$. Let $2^{x}=y$. Then the equation becomes $y^{3}-5 y^{2}+y+10=0$. Divide the left side of the equation by $y-2$:

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0434.jpg?height=145&width=477&top_left_y=499&top_left_x=153)

$$
\begin{aligned}
& -3 y^{2}+y \\
& \begin{array}{r}
\frac{-3 y^{2}+6 y}{-5 y+10} \\
-\frac{-5 y+10}{0}
\end{array}
\end{aligned}
$$

The equation can be represented as $(y-2)\left(y^{2}-3 y-5\right)=0$, from which $y_{1}=2, y_{2,3}=\frac{3 \pm \sqrt{29}}{2}$. We obtain: $2^{x}=2 \Rightarrow x_{1}=1 ; 2^{x}=\frac{3-\sqrt{29}}{2}<0$ (no solutions); $2^{x}=\frac{3+\sqrt{29}}{2} \Rightarrow x_{3}=\log _{2} \frac{3+\sqrt{29}}{2}=\log _{2}(3+\sqrt{29})-1$.

Answer: $1 ; \log _{2}(3+\sqrt{29})-1$.