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Suppose that an epidemic of chicken pox hits a private elementary school. Initially, $3 \%$ of the students have caught the disease. After four weeks $22 \%$ have caught the disease. At what point will $90 \%$ of the students have succumbed to chicken pox? (Assume logistic growth.)

After 10.3 weeks

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 7

Applications of Exponential and Logarithmic Functions

Missouri State University

Baylor University

Lectures

01:09

Logistic Growth A student …

02:30

$A$ medical researcher is…

02:45

The logistic growth functi…

01:27

Given that the initial infection of Chicken Pox at a school was 3% of students and that four weeks later it reached to 22% of students. We'd like to find. When will 90% of the students have succumbed to chicken pox. So starting with this, we need to solve for c. So this is going to be that first part of the first equation, but we're going to use the second piece at time zero to solve for C. So we're told that essentially at time zero or this initial time 3% of students were infected. Let's go ahead and plug this into that second formula. So we have 0.033% right equal to one divided by one plus C, Solving for C. right here we see that C is equal to 32.33. This gives us a sort of generic equation right here. So we have at time t this is equal to one divided by one plus 32.33 E. To the power of negative Katie. So this is just substituting our value for C. Into that first equation that we have over here. Now we're going to go ahead and use the second piece of information that was given to us, which was four weeks later, 20-22% of students were infected. So we're just going to plug in this information to our generic equation and blue right here. So we have 22% of students, so 0.22 is equal to one divided by one plus 32.33 E. To the power of negative Katie. So at this is that time for four weeks later. So we'll have negative four K. And what we need to do is just solve for case let's go ahead and clear these fractions which will give us 0.22 plus. Let's see. 7.1126. E. -4 K. is equal to one. Simplifying this. So it's subtract .22 from both sides gives us 7.1126. Each of the power of -4 K is equal to 0.78, Dividing both sides by 7.1126 leaves us with each of the power of -4 K. is equal to 0.10966. Taking the natural log of both sides. And then dividing by that -4 will give us our value for K. See that K is equal to 0.55- six. So that's another important piece of information. Now we're able to plug that back into this generic equation that we found up top. So coming back to here because we already found. See But now we have case we have That infection at time. T. is equal to one Divided by one plus 32.33 E. K. is negative so negative 0.55-6 T. All right. So now that we have this now we can finally solve for when will 90% of students have chickenpox? Because now we have every variable except for T. And that's what we're looking for anyways. So We C 90%. So that's 0.9. This is that infection level that we're looking for. It's equal to one divided by One plus 32.33. E. Really we're just plugging these values in that we know which was that .99.90. Now we just need to solve for T. So it's clear these fractions which leaves us with 0.9 plus 29.0978 To negative 0.55- 60 is equal to one. Let's go ahead and subtract .9 from both sides and then also divide by this 29.97 were just simplifying. Let's get this e piece all on its own. And now we can take the natural log of both sides and divide by that negative 055 to six or that K value. And that will give us Our time T. Which is equal to 10 to six weeks. It tells us that at about 10.3 weeks about 90% of that student population will have succumbed to chicken pox.

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