11. Let $a$ and $b$ be real numbers such that $a>b, 2^{a}+2^{b}=75$ and $2^{-a}+2^{-b}=12^{-1}$. Find the value of $2^{a-b}$.
11. Answer: 4.
$$
\frac{75}{12}=\left(2^{a}+2^{b}\right)\left(2^{-a}+2^{-b}\right)=2+2^{a-b}+2^{b-a} \text {. Then } 2^{a-b}+2^{b-a}=\frac{17}{4}=4+\frac{1}{4} \text {. }
$$

Since $a>b$, we obtain $2^{a-b}=4$.