Question

The function$$D(h)=5 e^{-0.4 h}$$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. What is the time between injections?

Short answer

2.29 hours

Step-by-step solution

- Set up the equation

We are given the function for the drug's concentration: \[ D(h) = 5 e^{-0.4 h} \] We need to find the time, \( h \), when the concentration, \( D \), reaches 2 milligrams. Set \( D(h) = 2 \). Thus, set up the equation: \[ 2 = 5 e^{-0.4 h} \]

- Solve for \( e^{-0.4 h} \)

Divide both sides of the equation by 5 to isolate the exponential term: \[ \frac{2}{5} = e^{-0.4 h} \] Simplifying, we get: \[ 0.4 = e^{-0.4 h} \]

- Take the natural logarithm of both sides

To solve for \( h \), take the natural logarithm (\( \ln \)) of both sides of the equation: \[ \ln\left(\frac{2}{5}\right) = \ln(e^{-0.4 h}) \] Using the fact that \( \ln(e^x) = x \), we get: \[ \ln\left(\frac{2}{5}\right) = -0.4 h \]

- Solve for \( h \)

Finally, isolate \( h \) by dividing both sides by -0.4: \[ h = \frac{\ln\left(\frac{2}{5}\right)}{-0.4} \] Use a calculator to find the numeric value of \( \ln\left(\frac{2}{5}\right) \): \[ \ln\left(\frac{2}{5}\right) \approx -0.916 \] Thus, \[ h = \frac{-0.916}{-0.4} \approx 2.29 \] So, the time between injections is approximately 2.29 hours.

Key concepts

natural logarithm

In mathematics, the natural logarithm, represented as \(ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is an important function in various fields, such as calculus and algebra, because it simplifies computations involving exponential growth or decay.
To understand natural logarithms better, think of them as the inverse operations of exponentiation. For instance, if you know the result of an exponential function, you can use natural logarithms to find the exponent. This is particularly useful in pharmacokinetics, where drug concentration follows an exponential decay.
In the context of the given problem, we used the natural logarithm to solve for \(h\). By applying natural logarithms, we were able to invert the exponential function \(e^{-0.4h}\) and solve for the variable.

solving exponential equations

Solving exponential equations involves finding the variable in the exponent. The general form of such an equation can be represented as \(a^x = b\), where \(a\) is a constant and \(x\) is the exponent to be determined. In pharmacokinetics, equations often involve exponential decay or growth.
To solve these types of equations, follow these steps:

• Isolate the exponential term.
• Apply the natural logarithm (ln\bin of both sides.
• Use properties of logarithms to simplify the equation.
• Solve for the variable in the exponent.

In our problem, we started with the equation \(2 = 5e^{-0.4h}\). By isolating the exponential term (\(e^{-0.4h}\)), taking the natural logarithm, and then solving for \(h\), we found the time at which the drug concentration reaches 2 milligrams.

pharmacokinetics

Pharmacokinetics is a branch of pharmacology dedicated to understanding how drugs move through the body. It deals with the absorption, distribution, metabolism, and excretion of medications. One of the key concepts in pharmacokinetics is the drug concentration in the bloodstream over time.
Mathematical functions, often exponential in nature, are used to model drug levels. These functions help determine optimal dosing schedules to maintain effective drug levels while avoiding toxicity.
In our example, the function \(D(h) = 5e^{-0.4h}\) models the decay of drug concentration over time after administration. By solving for \(h\), we can predict when the drug level will fall to a specific concentration, hence knowing when the next dose is required.

drug concentration model

A drug concentration model describes how the concentration of a drug in the bloodstream changes over time. These models are essential for determining dosing intervals, ensuring therapeutic effectiveness and patient safety.
For the given problem, the function \(D(h) = 5e^{-0.4h}\) is an exponential decay model. Here, \(D(h)\) represents the drug concentration in milligrams, and \(h\) represents the time in hours. The constant \(5\) is the initial concentration when the drug is first administered, and the term \(e^{-0.4h}\) shows the rapidity of drug elimination from the body.
By understanding this model, we can calculate the specific time when the drug concentration will decrease to a predefined level, requiring the next dose. This ensures that the patient maintains the correct therapeutic dose.