The exposure volume EV of a photograph can be calculated from the time, t, in seconds that the shutter was open and the value of the f-stop,f, the picture was taken at using by using the logarithmic equation:
EV=logbase2(f^2/t)
a. use the properties of logarithms to expand the formula for EV:
b. Keith takes a picture of a waterfall with a shutter speed of 1/60 second at f/5.657. he likes the exposure of his picture, but want to leave the shutter open four times as long to better capture the motion of the water. What f-stop should he use on his next picture.
For b I somehow got that 169 is my f- stop but Im pretty sure that is way off because I know about photography and theres no such thing as an f stop at 169. Anyway I try to solve it I keep getting 169 though!
Help would greatly be appreaciated with this, thanks. Please show all work too so I know where my mistake is.
+129839 EV=logbase2(f^2/t)
EV = log2 f^2 - log2 t
EV = 2 log2f - log2 t
EV = 2 log2 (5.657 ) - log2 (1/60) ≈ 10.90696
So
10.90696 = 2log2 (f) - log2 (4/60)
10.90696 + log2(4/60) = 2 log2 f divide by 2
[ 10.90696 + log2(4/60) ]/ 2 = log2 f
This says that f = 2^ ([ 10.90696 + log2(4/60) ]/ 2) ≈ 11.3140
[ I hope this is a valid f stop !!! ]
