Hello,
Imagine a class C amplifier with 5V supply and 2W output.
Assuming 1V saturation at the drain or collector, you have a swing of 4V. To get 2W into a load, the drain or collector has to see R = 4^2/(2*2) = 4 Ohms. Of course you need a parallel inductive component also, but I regard this for now.
A 100 mW device with same swing needs 80 Ohms.
To have a broadband match from 80 ohms to 50 Ohms you need fewer components with respect to the matching from 4 ohms to 50 Ohms.