Use two transistors, one to convert the logic signal to 12V and then another to switch the 12v
simulate this circuit – Schematic created using CircuitLab
Seeing as I want to use the two transistors as saturated switches I need to keep the collector current to base current ratio small, IE less than 10 (but for small transistors (which typically have larger gains than big transistors) like these less aiming for less than 20 would probably be just fine)
Assuming the base emitter voltage is about 0.7v with Q1 on (conducting fully) R1 passes about 11.3mA, R3 accounts for 0.07mA and the rest (call it 11.2 mA comes from, Q2 base. with a 100mA load on Q2 collector that's a current ratio of about 9:1. Thus the collector-emitter voltage drop will be small and the power will flow to the load instead of heating up the transistor.
Q1 needs to accept a little under 12mA (I'm rounding up because two significant figures are plenty here I'm aiming for a ballpark, not a bull's eye)
so that means about 1.2mA into the base. from a 5V logic signal that's be about 3K, I guessed 4.7K which means about 0.9mA which is less than I wanted, but still plenty. At 10 mA through the collector a 2N3904 has a current gain of more than 50 so asking it to give a gain of 13 is not going to be a problem.