jmc

Since $f(7)=50$, we have $f^{-1}(50)=7$. Similarly, $f(3)=10$ and $f(5)=26$, so $f^{-1}(10)=3$ and $f^{-1}(26)=5$. Therefore, $$\begin{array}{rl}{f}^{-1}\left(f^{-1}(50)\times f^{-1}(10)+f^{-1}(26)\right)& =f^{-1}(7\times 3+5)\\ & =f^{-1}(26)=\boxed{5}.\end{array}$$