Chapter 5.4, Problem 123AYU

Drug Medication The function

$D\left(\text{h }\right)=\text{ 5}{e}^{-0.4h}$

can be used to find the number of milligrams $D$ of a certain drug that is in a patient's bloodstream $h$ hours after the drug was administered, When the number of milligrams reaches

$2$, the drug is to beadministered again. What is the time between injections?

To determine

The time between injections, if the number of milligrams reaches $2$, the drug is to be administered again, where the function $D\left(h\right)=5e^{-0.4h}$ can be used to find the number of milligrams $D$ of a certain drug that is in a patient’s bloodstream $h$ hours after the drug was administered and when the number of milligrams reaches $2$, the drug is to be administered again.

Answer to Problem 123AYU

Solution:

The time between injections is approximately $2$ hours.

Explanation of Solution

Given information:

The function $D\left(h\right)=5e^{-0.4h}$ can be used to find the number of milligrams $D$ of a certain drug that is in a patient’s bloodstream $h$ hours after the drug was administered, when the number of milligrams reaches $2$, the drug is to be administered again.

Explanation:

The number of milligrams $D$ of a certain drug that is in a patient’s bloodstream $h$ hours after the drug was administered.

Here, $D\left(h\right)=5e^{-0.4h}$

If the number of milligrams reaches $2$, the drug is to be administered again.

Thus, $2=5e^{-0.4h}$

Divide both sides by $5$,

$\Rightarrow \frac{2}{5}=e^{-0.4h}$

$\Rightarrow 0.4=e^{-0.4h}$

Taking natural log on both sides,

$\Rightarrow \ln \left(0.4\right)=\ln \left(e^{-0.4h}\right)$

By using $\ln \left(a^b\right)=b\ln a$,

$\Rightarrow \ln \left(0.4\right)=-0.4h\ln \left(e\right)$

$\Rightarrow \ln \left(0.4\right)=-0.4h$

Divide both sides by $-0.4$

$\Rightarrow \frac{\ln \left(0.4\right)}{-0.4}=h$

$\Rightarrow h=2.29072\approx 2$

Therefore, the time between injections is $2.29072\approx 2$ hours.