Question

Algebra 2B

Log Functions in the Real World

Directions: Disease can spread quickly without use of universal precautions. Suppose the spread of a direct contact disease in a stadium is modeled by the exponential equation P(t) = 10,000/(1 +e^3-t) where P(t) is the total number of people infected after t hours. (Use the estimate for e (2.718) or the graphing calculator for e in your calculations.)


1. Estimate the initial number of people infected with the disease. Show how you found your answer.

2. Assuming the disease does not present symptoms for 24 hours, how many people will have been infected after 3 hours? Show how you found your answer.

3. What is the maximum number of people who can become infected? (Note: e(3-t) will approach 0 for very large values of t).

4. Explain why your answer for Question #3 is the maximum.

5. The stadium needs to warn its guests about a rapid disease spread if it affects over 800 people.

6. After how many minutes should the stadium inform its guests of the disease? Show how you found your answer.

Answer 1

The spread of the disease is an illustration of exponential equations, where the variables are exponents.

The initial number of people affected

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

At the initial time, the value of t is 0.

So, we have:

$P(0) = \frac{10000}{1 +e^{3-0}}$

$P(0) = \frac{10000}{1 +e^{3}}$

Estimate the quotient

$P(0) = 474$

Hence, the number of people infected initially is 474

The infected people after three hours

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

After three hours, the value of t is 3.

So, we have:

$P(3) = \frac{10000}{1 +e^{3-3}}$

$P(3) = \frac{10000}{1 +e^0}$

Estimate the quotient

$P(3) = 5000$

Hence, the number of people infected after three hours is 5000

The maximum number of people infected

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

As t approaches infinity, $e^{3-t}$ approaches 0

So, we have:

$P_{Max} = \frac{10000}{1 +0}$

Estimate the quotient

$P_{Max} = 10000$

Hence, the maximum number of people infected is 10000

The maximum number of people infected is 10000 because the denominator $1 + e^{3-t}$ cannot exceed 1

Time to warn its spread to over 800 people

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

When if affects 800 people, we have:

$\frac{10000}{1 +e^{3-t}} > 800$

Divide both sides by 10000

$\frac{1}{1 +e^{3-t}} > 0.08$

Take the reciprocal of both sides

$1 +e^{3-t} > 12.5$

Subtract 1 from both sides

$e^{3-t} > 11.5$

Take the natural logarithm of both sides

$3-t > \ln(11.5)$

Solve for t

$t < 3 - \ln(11.5)$

$t < 0.558$

Convert to minutes

$t < 0.558 * 60$

$t < 33.5$

Hence, the stadium should inform the guests before 33.5 minutes

Read more about exponential equations at:

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