Question 23: If real numbers $a$, $b$, $c$ satisfy $2^{a}+4^{b}=2^{c}, 4^{a}+2^{b}=4^{c}$, find the minimum value of $c$.


Question 23, Answer: According to the condition, we have $\left(2^{c}-4^{b}\right)^{2}+2^{b}=4^{c}$, rearranging which we get $2^{c+1}=\frac{2^{b}+4^{2 b}}{4^{b}}=\frac{1}{2^{b}}+2^{2 b}=$ $\frac{1}{2^{b+1}}+\frac{1}{2^{b+1}}+2^{2 b} \geq 3 \sqrt[3]{\frac{1}{2^{b+1}} \cdot \frac{1}{2^{b+1}} \cdot 2^{2 b}}=\frac{3 \sqrt[3]{2}}{2}$, thus we know $c \geq \log _{2} \frac{3 \sqrt[3]{2}}{2}-1=\log _{2} 3-\frac{5}{3}$. When $b=-\frac{1}{3}$, $a=-\frac{5}{3}$, we have $c=\log _{2} 3-\frac{5}{3}$.