1. ## help exponential

A person drinks alcohol at a party. After her last drink, the alcohol level of her blood soon reaches a maximum of 0.28 milligram alcohol per milliliter of blood. If the half-life of alcohol in her blood is 2 hours, how long must she wait before driving at the legal limit of 0.08 milligram alcohol per milliliter of blood?

2. Originally Posted by getnaphd
A person drinks alcohol at a party. After her last drink, the alcohol level of her blood soon reaches a maximum of 0.28 milligram alcohol per milliliter of blood. If the half-life of alcohol in her blood is 2 hours, how long must she wait before driving at the legal limit of 0.08 milligram alcohol per milliliter of blood?
the half life is the time required for half the original amount of alcohol to remain.

we can do a rough estimate this way:

now she has 0.28 mg of alcohol in her blood, after 2 hours she will have (1/2)0.28 = 0.14 mg. after another 2 hours, she will have (1/2)0.14 = 0.07 mg. This is under the limit, so she needs to wait for just under 4 hours before driving.

if we want to be more specific, we have to make an equation.

Let $\displaystyle A(t)$ be the amount of alcohol per mL of blood remaining after time $\displaystyle t$
Let $\displaystyle A_0$ be the initial amount of alcohol in her blood

After each half-life period, we will have some multiple of 1/2 of the original amount remaining. so if we start with $\displaystyle A_0$ after two hours, we have $\displaystyle \frac {1}{2} A_0$ after two more hours, we have $\displaystyle \left( \frac {1}{2} \right)^2 A_0$, so we can come up with a general formula. since for half-life periods, $\displaystyle A(t)$ is some multiple of 1/2 times $\displaystyle A_0$, we get:

$\displaystyle A(t) = A_0 \left( \frac {1}{2} \right)^{kt} = A_0 \cdot 2^{-kt}$

Now, when $\displaystyle t = 2$ (which is the half-life), we have half the original amount remaining, so $\displaystyle A(2) = \frac {1}{2} A_0$, so plug these values in.

$\displaystyle \frac {1}{2} A_0 = A_0 \cdot 2^{-2k}$

$\displaystyle \Rightarrow \frac {1}{2} = 2^{-2k}$

$\displaystyle \Rightarrow 2^{-1} = 2^{-2k}$

equating coefficients we get:

$\displaystyle -1 = -2k$

$\displaystyle \Rightarrow k = \frac {1}{2}$

So finally, our formula is: $\displaystyle \boxed { A(t) = A_0 \cdot 2^{-t/2} }$

Here, $\displaystyle A_0 = 0.28$

We want $\displaystyle t$ when $\displaystyle A(t) = 0.08$, so we get:

$\displaystyle 0.08 = 0.28 \cdot 2^{-t/2}$

$\displaystyle \Rightarrow 0.285714285 = 2^{-t/2}$

$\displaystyle \Rightarrow \log_2 0.285714285 = \log_2 2^{-t/2}$

$\displaystyle \Rightarrow - \frac {t}{2} = \log_2 0.285714285$

$\displaystyle \Rightarrow t = -2 \log_2 0.285714285... \approx 3.615$ hours

3. Originally Posted by getnaphd
A person drinks alcohol at a party. After her last drink, the alcohol level of her blood soon reaches a maximum of 0.28 milligram alcohol per milliliter of blood. If the half-life of alcohol in her blood is 2 hours, how long must she wait before driving at the legal limit of 0.08 milligram alcohol per milliliter of blood?
If the half life is $\displaystyle t_{hl}$ then the concentration $\displaystyle c(t)$ at time $\displaystyle t$ after the concentration was $\displaystyle c_0$ is:

$\displaystyle c(t)=c_0 ~2^{-t/t_{hl}}$

Now when $\displaystyle c(t)=0.08 \mbox{ mg/ml}$ we have:

$\displaystyle 0.08=0.28~2^{-t/2}$

where time is measured in hours.

Take logs:

$\displaystyle \log(0.08) = \log(0.28) - (t/2) \log(2)$

which we can solve for $\displaystyle t$:

$\displaystyle t=2~\frac{\log(0.08/0.28)}{\log(2)}\approx 3.615 \mbox{ hours}$

RonL