How does the current limiting resistor for an LED affect current and voltage drops?
The LED forward voltage drop will remain (roughly) the same, but the current can change, so the calculation becomes (same equation solving for I):

$$I_{LED} = {(V_s - V_f)\over{R}}$$

So for a 3V \${V_f}\$ and a 5V supply, the \$100\Omega\$ resistor would give \${(5V - 3V)\over{100 \Omega }} = 20 mA\$.

So if you know what current you want, just plug the values in, e.g. for 10mA:

$$R = {(5V - 3V)\over{0.01 A}} = {200 \Omega}$$

Basically, the fact that the supply and the LED forward voltage can be relied upon to be pretty static, means that whatever value resistor you put in will also have a static voltage across it (e.g ~2V in this case), so it just leaves you to find out that voltage and select a resistance value according to the current you want. 

Below is the V-I curve of a diode (from the wiki LED page), notice the current sharply rises (exponentially) but voltage stays roughly the same when the "on" voltage is reached. 



For more accurate control of the current you would use a constant current, which is what most LED driver ICs provide.