Question

the drug valium is eliminated from the bloodstream exponentially with a half - life of 36 hours. suppose that a patient receives an initial dose of 30 milligrams of valium at midnight.
a. how much valium is in the patients blood at noon on the first day?
b. estimate when the valium concentration will reach 45% of its initial level.
a. how much valium is in the patients blood at noon on the first day?
there is approximately mg of valium in the patients blood at noon on the first day.
(round to the nearest tenth as needed.)

Explanation:

Step1: Determine the exponential - decay formula

The general formula for exponential decay is $A = A_0(\frac{1}{2})^{\frac{t}{h}}$, where $A_0$ is the initial amount, $t$ is the time elapsed, and $h$ is the half - life. Here, $A_0 = 30$ mg, $h = 36$ hours.

Step2: Calculate the time elapsed for part a

The patient takes the dose at midnight and we want to find the amount at noon on the first day. The time elapsed $t = 12$ hours.
Substitute $A_0 = 30$, $t = 12$, and $h = 36$ into the formula: $A = 30(\frac{1}{2})^{\frac{12}{36}}=30(\frac{1}{2})^{\frac{1}{3}}$.
We know that $(\frac{1}{2})^{\frac{1}{3}}=\sqrt[3]{\frac{1}{2}}\approx0.7937$. So $A = 30\times0.7937\approx23.8$ mg.

Step3: Calculate the time for part b

We want to find $t$ when $A = 0.45A_0$. Substitute into the formula $A = A_0(\frac{1}{2})^{\frac{t}{h}}$:
$0.45A_0=A_0(\frac{1}{2})^{\frac{t}{36}}$.
Since $A_0
eq0$, we can divide both sides by $A_0$ to get $0.45 = (\frac{1}{2})^{\frac{t}{36}}$.
Take the natural logarithm of both sides: $\ln(0.45)=\ln((\frac{1}{2})^{\frac{t}{36}})$.
Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.45)=\frac{t}{36}\ln(\frac{1}{2})$.
We know that $\ln(0.45)\approx - 0.7985$ and $\ln(\frac{1}{2})\approx - 0.6931$.
Then $t = 36\times\frac{\ln(0.45)}{\ln(\frac{1}{2})}=36\times\frac{- 0.7985}{- 0.6931}\approx41.5$ hours.

Answer:

a. 23.8 mg
b. Approximately 41.5 hours after the initial dose.