
Patrick B. answered 11/23/19
Math and computer tutor/teacher
G(0) = 156 <--- this is the initial value G0
G(210) = 2.4375
G(t) = 156*(1/2)^(t/h) where h is the half life
2.4375 = 156*(1/2)^(210/h)
0.015625 = (1/2)^(210/h)
takes the log of both sides, using change of base formula:
log_1/2(0.015625) = log(0.05625)/log(1/2) = 6
6 = 210/h
6h = 210
h = 210/6 = 35
the half life is 35 minutes
G(t) = 156*(1/2)^(t/35)
there will be 78 grams left after 35 minutes.