Question 1096997: The half-life of zulus is 14 days and they decay exponentially. If Angela begins with 20 zulus, how long will it take until only 5 remain? Found 3 solutions by jorel1380, ikleyn, greenestamps:Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website! The formula for exponential decay here is P(t)=P(0)e^-kt, where P(t) is the resultant population,P(0)is the initial population, k is the decay constant, and t is the time. So, in the case of half-life, P(t)/P(0)=.5, and t here is 14 days. So:
.5=e^-14k
ln 0.5=ln e^-14k=-14k ln e=-14k
k=0.04951051289713895067265943724701
For there to be 5 zulus left, we have:
5/20=e^-0.04951051289713895067265943724701t
.25=e^-0.04951051289713895067265943724701t
ln 0.25=ln e^-0.04951051289713895067265943724701t=-0.04951051289713895067265943724701t (ln e)=-0.04951051289713895067265943724701t
t=ln 0.25/-0.04951051289713895067265943724701=28 days
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In 14 days only half of initial 20 zulus will remain, i.e. 10 zulus will remain
In the next 14 says only half of these 10 zulus will remain, i.e. 5 zulus will remain.
Answer. In 14 + 14 = 28 days.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website! Tutor jorell380 is almost certainly a scientist more than a mathematician; scientists like to work exponential functions using e as the base. For me, as a mathematician, working half life problems using powers of 1/2 makes far more sense.
For problems with half lives, the exponential decay function is
y is the amount remaining; A is the initial amount; and x is the number of half lives.
For your problem, we have 20 zulus to start and 5 at the end. So the number of half lives is found by solving the following equation for x:
We don't need any logarithms there! 1/4 is (1/2)^2; it takes 2 half lives for the number of zulus to drop from 20 to 5.
The half life is 14 days, so 2 half lives is 28 days.