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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.Biologists consider a species of animal or plant to be endangered if it is expected to become extinct within20 years. If a certain species of wildlife is counted to have 1147 members at the present time, and the population has been steadily declining exponentially at an annual rate averaging $39 \%$ over the past 7 years, do you think the species is endangered? Explain your answer.

Yes, $y(20)<1$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

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So we're told that something is considered in danger if after 20 years, the population is expected to be extinct or within 20 years. So we're told that some wildlife has 1147 members at the current time, and population is following a steady decline exponentially, and it's been averaging a rate of 39% over the past seven years. We want to determine if this species is going to be endangered and we're gonna explain your answer. So here I write endangered if p of 20 is less than one. So if we have some population function pft Um, since they tell us it's exponential decline, we know our equation should look something like this to start. So P of t is equal to p, not times e to the negative Katie, and this is because it's declined. We have that negative there. And so we say, strictly less than one, as opposed to equal to zero, because exponential can ever equal to zero, at least with this model here. So that's why we're looking for less than one now. Let's go ahead and figure out what to plug it. Soapy. Not is supposed to be our initial population. And we're told initially it is 1147. So this is gonna be peanuts. So it's got unplugged batted, and this is just from the equations that they give us in the chapter. So we don't need to work harder than we need to, and we just plug everything in that they tell us we can use now. This next part, it says every year for the past seven years has been declining around 39%. So that means, after one year, soapy of one should be equal to zero point 61 of 11 47 which is p not so. We'll just go ahead and set this equal to our function over here, and we'll have e to the negative K, and we need to plug it one for teeth. Now. The reason why I wrote it out as six point one 4.61 as opposed to multiply and those together is notice that when we would divide by our initial population, those just can't stop at this point, we would want to take the natural log on each side and then multiply by negative one and that's going to give us. Kay is even too negative. Natural log of 0.61 So we have Kay. Now let's go ahead and plug that into our original equation. So we're gonna get that population after tea years is going to be a 1147 e to the So the negatives will counts out. So it should just be natural of 0.61 times teeth. Now we want to check to see if P of 20 is less than one. So let's plug in 20. And actually, I'll be a little bit lazy and we'll just replace thes tease right here with 20. And so let's see what we get after we plug. Everything is so this is going to be approximately 1107 and in the numerator 20 times. The natural log a point 61 is he to the negative 9.88 about and now exponentially ating. That gives us something around five times 10 to the negative fifth and then multiply that by 1147. So this is going to give us the population of about 0.5 So we did. We can't have a fraction of an organism and it's considered to be alive. So this is less than one which meets our criteria of being endangered. So we would just say something like, yes, since P of 20 is equal to 0.5 which is less than one.

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