Question

the concentration of a drug in the body decreases exponentially after a dosage is given. in one clinical study, adult subjects averaged 12 micrograms/milliliter (mcg/ml) of the drug in their blood plasma 1 hr after a 1000 - mg dosage and 5 micrograms/milliliter 7 hr after dosage. assume the concentration decreases according to the exponential decay model.
a) find the value k, and write an equation for an exponential function that can be used to predict the concentration of the drug, in micrograms/milliliter, t hours after a 1000 - mg dosage.
b) estimate the concentration of the drug 5 hr after a 1000 - mg dosage.
c) to relieve a fever, the concentration of the drug should go no lower than 4 mcg/ml. after how many hours will a 1000 - mg dosage drop to that level?
a) k = 0.146 (round to three decimal places as needed.)
choose the correct equation below
a. p(t)=15900e^(-0.146t)
b. p(t)=11900e^(-0.146t)
c. p(t)=15900e^(-0.730t)
d. p(t)=13.9e^(-0.268t)
e. p(t)=12e^(-0.146t)
f. p(t)=13.9e^(-0.146t)
g. p(t)=12e^(-0.029t)
h. p(t)=13.9e^(-0.730t)

Explanation:

Step1: Recall exponential - decay formula

The exponential - decay formula is $P(t)=P_0e^{-kt}$, where $P(t)$ is the concentration of the drug at time $t$, $P_0$ is the initial concentration, $k$ is the decay constant, and $t$ is the time. We know that at $t = 1$, $P(1)=12$ and at $t = 7$, $P(7)=5$. Substituting into the formula: $P(1)=P_0e^{-k\times1}=12$ and $P(7)=P_0e^{-k\times7}=5$. Dividing the second equation by the first equation: $\frac{P_0e^{-7k}}{P_0e^{-k}}=\frac{5}{12}$. Using the property of exponents $\frac{e^a}{e^b}=e^{a - b}$, we get $e^{-6k}=\frac{5}{12}$. Taking the natural - logarithm of both sides: $\ln(e^{-6k})=\ln(\frac{5}{12})$. Since $\ln(e^x)=x$, we have $-6k=\ln(\frac{5}{12})$. Solving for $k$, $k=-\frac{\ln(\frac{5}{12})}{6}\approx0.146$.

Step2: Find the exponential function

We know that at $t = 1$, $P(1) = 12$. Using the formula $P(t)=P_0e^{-kt}$, when $t = 1$ and $k = 0.146$, $12=P_0e^{-0.146\times1}$. Solving for $P_0$, $P_0=\frac{12}{e^{-0.146}}\approx13.9$. So the exponential function is $P(t)=13.9e^{-0.146t}$.

Step3: Estimate the concentration at $t = 5$

Substitute $t = 5$ into the function $P(t)=13.9e^{-0.146t}$. Then $P(5)=13.9e^{-0.146\times5}=13.9e^{-0.73}\approx6.6$ micrograms/milliliter.

Step4: Find the time when $P(t)=4$

Set $P(t)=4$ in the function $P(t)=13.9e^{-0.146t}$. So $4 = 13.9e^{-0.146t}$. First, divide both sides by 13.9: $\frac{4}{13.9}=e^{-0.146t}$. Then take the natural - logarithm of both sides: $\ln(\frac{4}{13.9})=-0.146t$. Solving for $t$, $t=-\frac{\ln(\frac{4}{13.9})}{0.146}\approx8.7$ hours.

Answer:

a) The correct equation is D. $P(t)=13.9e^{-0.146t}$
b) Approximately $6.6$ micrograms/milliliter
c) Approximately $8.7$ hours