9. (16 points) If real numbers $a$, $b$, $c$ satisfy
$$
2^{a}+4^{b}=2^{c}, 4^{a}+2^{b}=4^{c} \text {, }
$$

find the minimum value of $c$.
9. Let $2^{a}, 2^{b}, 2^{c}$ be denoted as $x, y, z$ respectively.
Then $x, y, z > 0$.
$$
\begin{array}{l}
\text { Hence } x+y^{2}=z, x^{2}+y=z^{2} \\
\Rightarrow z^{2}-y=x^{2}=\left(z-y^{2}\right)^{2}=z^{2}-2 y^{2} z+y^{4} \\
\Rightarrow z=\frac{y^{4}+y}{2 y^{2}}=\frac{1}{4}\left(2 y^{2}+\frac{1}{y}+\frac{1}{y}\right) \\
\quad \geqslant \frac{1}{4} \times 3 \sqrt[3]{2 y^{2} \cdot \frac{1}{y} \cdot \frac{1}{y}}=\frac{3 \sqrt[3]{2}}{4} .
\end{array}
$$

When $2 y^{2}=\frac{1}{y}$, i.e., $y=\frac{1}{\sqrt[3]{2}}$, then $z_{\text {min }}=\frac{3 \sqrt[3]{2}}{4}$.
At this point, the corresponding $x_{\text {min }}=\frac{\sqrt[3]{2}}{4}$, which meets the requirements. Since $c=\log _{2} z$, the minimum value of $c$ is $\log _{2}\left(\frac{3 \sqrt[3]{2}}{4}\right)=\log _{2} 3-\frac{5}{3}$.