Question

the logistic growth function $p(x)=\frac{90}{1 + 271e^{-0.122x}}$ models the percentage, $p(x)$, of americans who are $x$ years old and have some coronary heart disease. use this function to find the age at which the percentage of americans who have some coronary heart disease is 85%.
what is the age at which 85% of americans have some coronary heart disease?
$\square$ years old (round to the nearest integer.)

Explanation:

Step1: Substitute \(P(x) = 85\) into the function

We have the equation \(85=\frac{90}{1 + 271e^{-0.122x}}\).
First, cross - multiply: \(85(1 + 271e^{-0.122x})=90\).
Then, distribute: \(85+85\times271e^{-0.122x}=90\).
Subtract 85 from both sides: \(85\times271e^{-0.122x}=90 - 85=5\).
So, \(e^{-0.122x}=\frac{5}{85\times271}=\frac{5}{23035}\approx0.000217\).

Step2: Take the natural logarithm of both sides

Using the property \(\ln(e^{a})=a\), if \(y = e^{-0.122x}\), then \(\ln(y)=-0.122x\).
\(\ln(e^{-0.122x})=\ln(0.000217)\).
Since \(\ln(e^{-0.122x})=-0.122x\), we have \(-0.122x=\ln(0.000217)\).
We know that \(\ln(0.000217)\approx - 8.33\).

Step3: Solve for \(x\)

\(x=\frac{\ln(0.000217)}{- 0.122}\).
Substitute \(\ln(0.000217)\approx - 8.33\) into the formula: \(x=\frac{-8.33}{-0.122}\approx68.3\).

Answer:

\(68\)