I got a bit confused when I read some of the replies which mentioned a 3.3V supply and 1k resistor.
Yes, with a potential divider after the 3.3V, making the base voltage 3.1V and a 10M mulitmeter loading the collector, the base-emitter drop will be lower, giving about 3.8V.
The lower the base current, the lower the base emitter voltage, so in this case it drops just over 0.3V, rather than the usual 0.6V. I've added some arrows which shows the reading you'd get on multimeter.
The key to understanding this is all voltages are relative to one another, with ground or 0V, the common point in the circuit from where all voltages are measured from.
V
B = 3.15V
V
BE = 0.34V
V
E = V
B = V
BE = 3.15 - 0.34 = 2.81V