problem: Given that $f(2)=5$ and $f^{-1}(x+4)=2f^{-1}(x)+1$ for all $x$, find $f^{-1}(17)$.
solution: Note that $f(2)=5$ implies $f^{-1}(5)=2$. Applying $f^{-1}(x+4)=2f^{-1}(x)+1$ repeatedly, we have: \begin{align*}
  f^{-1}(5)&=2 \\
  \Rightarrow \quad f^{-1}(9)&=2f^{-1}(5)+1=5 \\
  \Rightarrow \quad f^{-1}(13)&=2f^{-1}(9)+1=11 \\
  \Rightarrow \quad f^{-1}(17)&=2f^{-1}(13)+1=23.
  \end{align*}So $f^{-1}(17)=\boxed{23}$.