When a certain drug enters the blood stream, its potency decreases exponentially with a half life of 6 hours. Suppose the initial amount of the drug present is A (sub 0). How much of the drug will be present after each number of hours?
a. 6 b. 24 c. t
Thank you so much for your help!
When a certain drug enters the blood stream, its potency decreases exponentially with a half life of 6 hours. Suppose the initial amount of the drug present is A (sub 0). How much of the drug will be present after each number of hours?
a. 6 b. 24 c. t
Amount=Ao* (1/2)^t/6 is one way to solve the question. Another is
Amount=Ao*e^(-.692t/6)
where t is time.
no clue
Let A be the amount left after time t, thus we have A(t). We're told this is a negative exponential of the form Ke^(-rt) where r is the rate of decrease and t is the time. K is an arbitrary constant that needs to be determined. So we have
(1) A(t)=Ke^(-rt)=Ke^(-rt)
When t=0, K = the amount entered. When t=6, A=(1/2)K=Ke^(-r*6), so
1/2=e^(-r*6) or ln(1/2)=-6r or
r=(-1/6)*ln(1/2)
If you use this in (1) you'll get
A(t)=Ke^(-(-1/6)*ln(1/2)
*t) or
(2) A(t)=K*(1/2)^((1/6)*t)
Now answer the questions. Always test t=0 and the number that gives the first half-life. For t=0, A(t)=K and for t=6, A(t)=(1/2)K
This is a gerneral way to solve exponential problems.
I'll let you finish it.
To solve this problem, we need to use the concept of exponential decay. The formula for exponential decay is A(t) = Ao * (1/2)^(t/6), where A(t) is the amount of the drug present at time t, Ao is the initial amount of the drug, and t is the time in hours.
a. For t = 6 hours, we substitute t = 6 into the formula:
A(6) = Ao * (1/2)^(6/6)
A(6) = Ao * (1/2)^1
A(6) = Ao * (1/2)
A(6) = (1/2)Ao
b. For t = 24 hours,
A(24) = Ao * (1/2)^(24/6)
A(24) = Ao * (1/2)^4
A(24) = Ao * (1/16)
c. For t = t hours,
A(t) = Ao * (1/2)^(t/6)
Remember to substitute the value of Ao, the initial amount of the drug, to get the final result.
In summary:
a. After 6 hours, the amount of the drug present is (1/2)Ao.
b. After 24 hours, the amount of the drug present is (1/16)Ao.
c. After t hours, the amount of the drug present is Ao * (1/2)^(t/6).