Let f(x) = 16x + 3. Find the sum of all x that satisfy the equation f-1(x) = f((2x)-1).
$f(x)=y\implies 16x+3=y\implies x=\frac{y-3}{16}$ so $f^{-1}(x)=\frac{x-3}{16}$. The continuation is simple.
+30 
+600 
