$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$7.8 \quad 4^{\sqrt{x}}-9 \cdot 2^{\sqrt{x}-1}+2=0$.
7.8 Since $4^{\sqrt{x}}=2^{2 \sqrt{x}}$ and $2^{\sqrt{x}-1}=2^{\sqrt{x}} \cdot 2^{-1}=\frac{1}{2} \cdot 2^{\sqrt{x}}$, the given equation will take the form

$$
2^{2 \sqrt{x}}-\frac{9}{2} \cdot 2^{\sqrt{x}}+2=0
$$

Let's make the substitution $2^{\sqrt{x}}=y$, where $y>0$ due to the property of the exponential function $2^{0}$. Then we get the equation $y^{2}-\frac{9}{2} y+2=0$, the roots of which are $y_{1}=4$, $y_{2}=\frac{1}{2}$, both positive. From the equation $2^{\sqrt{x}}=4$ we have $2^{\sqrt{x}}=2^{2}, \sqrt{x}=2$, hence $x=4$. From the equation $2^{\sqrt{x}}=\frac{1}{2}$ we find $2^{\sqrt{x}}=2^{-1}$, hence $\sqrt{x}=-1$, which is impossible.

Thus, $x=4$.

Answer: $x=4$.