9. If real numbers $a, b, c$ satisfy $2^{a}+4^{b}=2^{c}, 4^{a}+2^{b}=4^{c}$, find the minimum value of $c$.

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Let $x=2^{a}, y=2^{b}, z=2^{c}$, then 
\[
\left\{\begin{array}{l}
x+y^{2}=z, \\ 
x^{2}+y=z^{2}
\end{array} \Rightarrow\left\{\begin{array}{l}
x=z-y^{2}, \\ 
x^{2}=z^{2}-y
\end{array}\right.\right.
\]
\[
\begin{array}{l}
\Rightarrow z^{2}+y^{4}+2 x y^{2}=z^{2}-y \Rightarrow 2 y^{2} z=y^{4}+y \Rightarrow 2 z=y^{2}+\frac{1}{y} \\
=y^{2}+\frac{1}{2 y}+\frac{1}{2 y} \geqslant \frac{3}{\sqrt[3]{4}}=\frac{3 \sqrt[3]{2}}{2} \Rightarrow z \geqslant \frac{3 \sqrt[3]{2}}{4},
\end{array}
\]

Equality holds when $y^{2}=\frac{1}{2 y} \Rightarrow y=\frac{1}{\sqrt[3]{2}} \Rightarrow x=z-y^{2}=\frac{3}{2 \sqrt[3]{4}}-\frac{1}{\sqrt[3]{4}}=\frac{\sqrt[3]{2}}{4}$.
Thus, $2^{c} \geqslant \frac{3 \sqrt[3]{2}}{4} \Rightarrow c \geqslant \log _{2} 3-\frac{5}{3}$. Therefore, the minimum value of $c$ is $\log _{2} 3-\frac{5}{3}$.