Question

(group quiz)
linear and exponential functions practice
) suppose a new drug, drug x, dissipates in the bloodstream at a continuous rate of −0.17. suppose joe ingests 100 mg of drug x (and all of it goes into his bl
rite the exponential function for the amount of drug x in joe’s bloodstream a
nt: we are given the continuous rate, so what form should we use?)
) determine the half-life of drug x in the bloodstream.
drug x if there is at least 1 mg of it

Explanation:

Part (a): Writing the Exponential Function

Step1: Recall continuous decay formula

The formula for continuous exponential decay is $A(t) = A_0 e^{kt}$, where $A_0$ is the initial amount, $k$ is the continuous rate, and $t$ is time. Here, $A_0 = 100$ mg, $k = -0.17$ (negative for decay).

Step2: Substitute values into formula

Substitute $A_0 = 100$ and $k = -0.17$ into the formula: $X(t) = 100 e^{-0.17t}$.

Step1: Define half - life condition

Half - life is the time $t$ when $X(t)=\frac{1}{2}A_0$. Since $A_0 = 100$, we want $X(t) = 50$. So we set up the equation $50=100 e^{-0.17t}$.

Step2: Solve for $t$

First, divide both sides by 100: $\frac{50}{100}=e^{-0.17t}$, which simplifies to $\frac{1}{2}=e^{-0.17t}$.
Then, take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{-0.17t})$.
Using the property $\ln(e^x)=x$, we get $\ln(\frac{1}{2})=- 0.17t$.
We know that $\ln(\frac{1}{2})=-\ln(2)$, so $-\ln(2)=-0.17t$.
Divide both sides by $- 0.17$: $t=\frac{\ln(2)}{0.17}$.
Calculate $\ln(2)\approx0.6931$, so $t=\frac{0.6931}{0.17}\approx4.08$ (rounded to two decimal places).

Answer:

$X(t) = 100 e^{-0.17t}$

Part (b): Determining the Half - Life