9. (This question is worth 16 points) If real numbers $a, b, c$ satisfy $2^{a}+4^{b}=2^{c}, 4^{a}+2^{b}=4^{c}$, find the minimum value of $c$.


Solution: Let $2^{a}, 2^{b}, 2^{c}$ be denoted as $x, y, z$, respectively, then $x, y, z > 0$.
From the conditions, we have $x + y^{2} = z, x^{2} + y = z^{2}$, so $z^{2} - y = x^{2} = (z - y^{2})^{2} = z^{2} - 2 y^{2} z + y^{4}$. 8 points
Therefore, combining the AM-GM inequality, we get
$$
z = \frac{y^{4} + y}{2 y^{2}} = \frac{1}{4} \left(2 y^{2} + \frac{1}{y} + \frac{1}{y}\right) \geq \frac{1}{4} \cdot 3 \sqrt[3]{2 y^{2} \cdot \frac{1}{y} \cdot \frac{1}{y}} = \frac{3}{4} \sqrt[3]{2}. 12 \text{ points}
$$

When $2 y^{2} = \frac{1}{y}$, i.e., $y = \frac{1}{\sqrt[3]{2}}$, the minimum value of $z$ is $\frac{3}{4} \sqrt[3]{2}$ (at this point, the corresponding $x$ value is $\frac{3}{4} \sqrt[3]{2}$, which meets the requirements).
Since $c = \log_{2} z$, the minimum value of $c$ is $\log_{2} \left(\frac{3}{4} \sqrt[3]{2}\right) = \log_{2} 3 - \frac{5}{3}$. 16 points