Assume that I'm taking the photo of a lamp in an outdoor cloudy day. The lamp itself has the luminance corresponding to 17EV. The environment has the illuminance corresponding to 14EV.
What will be the exposure value of the lamp?
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The math on two fstops is the square root of the sum of the two squares, but I don't know about EV.
If the lights were equal, and if they lighted the same overlapping area, twice the light is one stop additional. But you say the ambient is 3 stops down from the lamp, so it won't have much effect. This ambient adds less than 0.2 stop increase, not over 17.2EV.