A radioactive substance decays to the formula N=8e^-0.6t , where N is the number of milligrams present after t hours. Determine the initial amount of the substance to the nearest 10th of an hour, determine the half life of the substance, and determine the number of hours for 0.3 mg to remain.

Answer 1

"Determine the initial amount of the substance to the nearest 10th of an hour"
HUH, since when is an amount of something measured in units of time?? (I will just ignore that part of the question)

N = 8 e^(-6t)
initial --- t=0
N = 8 e^0 = 8 mg <----- the initial amount

for half-life
.5 = e^(-.6t)
take ln of both sides
ln .5 = -.6t ln e, but lne = 1
-.6t = ln .5
t = 1.155..
the half life is 1.155 hrs or 1 hour and appr 9 minutes

.3 = 8 e^(-.6t)
.0375 = e^(-.6t)
ln both sides
ln .0375 = -.6t
t = 5.47 hrs or 5 hours and appr 28 minutes

Answer 2

Omg you are amazing. You don't even know. I have been sitting here trying to figure this out for hours. Thank you so much! (Idk about the time thing, that's how it's written) THANK YOU!