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 = (round to three decimal places as needed.)

Explanation:

Step1: Recall exponential - decay formula

The exponential - decay model is given by $C(t)=C_0e^{-kt}$, where $C(t)$ is the concentration of the drug at time $t$, $C_0$ is the initial concentration, $k$ is the decay constant. We know that $C(1) = 12$ and $C(7)=5$.
Using $C(t)=C_0e^{-kt}$, when $t = 1$, $C(1)=C_0e^{-k}=12$; when $t = 7$, $C(7)=C_0e^{-7k}=5$.
Dividing the second equation by the first equation: $\frac{C_0e^{-7k}}{C_0e^{-k}}=\frac{5}{12}$.

Step2: Simplify the ratio

By the properties of exponents $\frac{e^{-7k}}{e^{-k}}=e^{-7k + k}=e^{-6k}$. So, $e^{-6k}=\frac{5}{12}$.
Take the natural - logarithm of both sides: $\ln(e^{-6k})=\ln(\frac{5}{12})$.
Since $\ln(e^{-6k})=-6k$, we have $-6k=\ln(\frac{5}{12})$.

Step3: Solve for $k$

$k=-\frac{1}{6}\ln(\frac{5}{12})=\frac{1}{6}\ln(\frac{12}{5})$.
Using a calculator, $k=\frac{\ln(12)-\ln(5)}{6}\approx\frac{2.4849 - 1.6094}{6}=\frac{0.8755}{6}\approx0.146$.
The exponential function is $C(t)=C_0e^{-0.146t}$. To find $C_0$, substitute $t = 1$ and $C(1)=12$ into $C(t)=C_0e^{-0.146t}$: $12 = C_0e^{-0.146}$, so $C_0=\frac{12}{e^{-0.146}}=12e^{0.146}\approx12\times1.1576 = 13.891$. The function is $C(t)=13.891e^{-0.146t}$.

Step4: Estimate the concentration at $t = 5$

Substitute $t = 5$ into $C(t)=13.891e^{-0.146t}$.
$C(5)=13.891e^{-0.146\times5}=13.891e^{-0.73}$.
$C(5)=13.891\times0.4827\approx6.70$.

Step5: Find the time when $C(t)=4$

Set $C(t)=4$ in $C(t)=13.891e^{-0.146t}$.
$4 = 13.891e^{-0.146t}$.
$e^{-0.146t}=\frac{4}{13.891}\approx0.288$.
Take the natural - logarithm of both sides: $\ln(e^{-0.146t})=\ln(0.288)$.
$-0.146t=\ln(0.288)$.
$t=-\frac{\ln(0.288)}{0.146}=\frac{-\ln(0.288)}{0.146}\approx\frac{-(-1.244)}{0.146}\approx8.52$.

Answer:

a) $k\approx0.146$
b) $C(t)=13.891e^{-0.146t}$
c) Approximately $6.70$ mcg/mL
d) Approximately $8.52$ hours