Problem 29 - College Algebra

Question

The logistic growth function $$P(x)=\frac{90}{1+271 e^{-0.122 x}}$$ models the percentage, \(P(x),\) of Americans who are \(x\) years old with some coronary heart disease. At what age is the percentage of some coronary heart -disease \(50 \% ?\)

Step-by-Step Solution

Answer

Approximately at the age of 43, 50% of Americans have some coronary heart disease.

Step 1: Setting up the equation

To start, you replace \(P(x)\) in the given equation with 50. The equation then becomes: \[50=\frac{90}{1+271 e^{-0.122 x}}\]

Step 2: Isolate expression containing \(x\)

You need to isolate the part of the equation containing \(x\). Start by multiplying both sides by the denominator of the right side of the equation, giving you: \[50 \cdot (1+271 e^{-0.122 x}) = 90.\] Now, subtract 50 from both sides: \[50 \cdot 271 e^{-0.122 x} = 90 - 50.\]

Step 3: Solve for \(x\)

Now you can start solving for \(x\). First, divide by 50*271 to get: \[e^{-0.122 x} = \frac{90 - 50}{50 \cdot 271}.\] To isolate \(x\), take the natural logarithm on both sides: \[-0.122x = \ln(\frac{90 - 50}{50 \cdot 271}).\] Now divide by -0.122 to get: \[x = \frac{\ln(\frac{40}{50 \cdot 271})}{-0.122}.\]

Step 4: Calculate \(x\)

Finally, plug the values into a calculator to get the value for \(x\): \[x \approx 42.6.\] This means that approximately 50% of Americans have some coronary heart disease at roughly 43 years old.

Key Concepts

Coronary Heart Disease Modeling

The logistic growth function is a powerful tool used in the modeling of various populations and phenomena, including medical conditions like coronary heart disease. In this context, the logistic growth function helps us understand how the percentage of individuals within a given age group may have coronary heart disease. This particular model takes the form: \[ P(x) = \frac{90}{1 + 271 e^{-0.122 x}} \] Here, \( P(x) \) represents the percentage of people aged \( x \) who are affected by coronary heart disease. The logistic model is particularly useful because it accounts for the growth or spread of a condition over time or age, reaching a maximum limit. This is relevant in medical modeling since, generally, not every individual will be affected, and the disease prevalence will plateau at a certain point. Such models help in understanding the likely age prevalence, providing insights to healthcare providers for better resource allocation and preventive strategies. Understanding the equation's parameters, such as the maximum percentage and the rate of increase, enables us to make predictions and work out when a specific percentage of the population might be affected.

Solving Exponential Equations

Exponential equations, like the one found in logistic growth models, can be tricky to solve, but following a structured approach makes it manageable. In our exercise, the equation derived from the logistic model was \[ 50 = \frac{90}{1+271 e^{-0.122 x}}. \] The goal here is to find the value of \( x \) (the age) where 50% of individuals have coronary heart disease. Solving such equations typically involves:

• Clearing the fraction by multiplying both sides by the denominator.
• Isolating the exponential term containing \( x \).
• Using logarithms to solve for \( x \).

In our case, isolating the exponential term is achieved by multiplying the whole equation with the denominator and then simplifying to gradually remove terms on one side of the equation. This process transforms the original logistic function into a simpler form, letting us use algebraic manipulation. The simplified equation allows us to apply logarithms and finally calculate the desired age \( x \) where the equation equals 50. Remember, practice with clear steps, and repeated application reinforces understanding.

Natural Logarithms

Natural logarithms are a fundamental concept in mathematics, especially useful in solving exponential equations. The natural logarithm is denoted as \( \ln \), and it is the inverse operation to exponentiation with base \( e \), where \( e \approx 2.718 \). When solving exponential equations like the one in our exercise, natural logarithms come into play as they help us "bring down" the variable exponent:\[ e^{-0.122 x} = \frac{40}{50 \times 271} \]To solve for \( x \), we use the natural logarithm on both sides: \[ -0.122x = \ln \left( \frac{40}{50 \times 271} \right) \]This step is crucial because it allows us to transform an exponential equation into a linear one, which is much simpler to solve. Dividing the result by \(-0.122\) gives us the age \( x \), at which the percentage reaches 50%. Natural logarithms are particularly favored in calculus and real-world applications for their property of converting multiplicative relationships into additive ones, thus simplifying complex computations.