15.10. If $f(x+y)=f(x) \cdot f(y)$ for all positive integers $x, y$ and $f(1)=2$, find $f(2007)$.
15.10. $2^{2007}$
By induction, it can be shown that $f(n)=[f(1)]^{n}=2^{n}$ for all positive integers $n$. Thus, $f(2007)=2^{2007}$.