Video: Predicting a Population That Grows According to a Logistic Model given the Parameters of the Model

Suppose a population grows according to a logistic model with an initial population of 1000 and a carrying capacity of 10000. If the population grows to 2500 after one year, what will the population be after another three years?

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Video Transcript

Suppose a population grows according to a logistic model with an initial population of 1000 and a carrying capacity of 10000. If the population grows to 2500 after one year, what will the population be after another three years?

We know that the general solution to the logistic model is given by 𝑃 of 𝑑 is equal to 𝐿 over one plus 𝐴𝑒 to the negative π‘˜π‘‘, where 𝐴 is equal to 𝐿 minus 𝑃 nought over 𝑃 nought. Here 𝐿 is the carrying capacity of the population, 𝑃 nought is the initial population, and π‘˜ is the growth rate of the population.

We’ve been given some of this information in the question. We’re told that the initial population, 𝑃 nought, is 1000. And we’re told that the carrying capacity, 𝐿, is 10000. But we haven’t been given the growth rate of the population. Instead, we’ve been given another pair of values for 𝑃 and 𝑑. We’re told that the population after one year is 2500. We’ll be able to combine this information with our values of 𝐿 and 𝑃 nought in order to determine the growth rate of the population.

First, we can work out the value of the constant 𝐴. It’s 𝐿 minus 𝑃 nought over 𝑃 nought, 10000 minus 1000 over 1000, which is 9000 over 1000, which is nine. Substituting 𝐿 and 𝐴 into our model then, we have that 𝑃 of 𝑑 is equal to 10000 over one plus nine 𝑒 to the power of negative π‘˜π‘‘.

Now we can use the population after one year in order to determine the value of π‘˜. Substituting 2500 for 𝑃 and one for 𝑑, we have 2500 equals 10000 over one plus nine 𝑒 to the negative π‘˜. To solve for π‘˜, we first multiply by one plus nine 𝑒 to the negative π‘˜ and then divide by 2500, giving one plus nine 𝑒 to the negative π‘˜ is equal to four. We can then subtract one and divide by nine, giving 𝑒 to the negative π‘˜ equals four minus one over nine, which is three-ninths or one-third.

We then take natural logarithms of each side, knowing that this will cancel out with the exponential on the left-hand side, to give negative π‘˜ equals the natural logarithm of one-third. We can then multiply by negative one to give π‘˜ equals negative the natural logarithm of one-third. And using laws of logarithms, this is equal to the natural logarithm of three. So we found the value of π‘˜, the growth rate of the population. Our model therefore becomes 𝑃 of 𝑑 equals 10000 over one plus nine 𝑒 to the negative 𝑑 multiplied by the natural logarithm of three.

Now we’re asked for the population after another three years, which means we’re looking for the population when 𝑑 is equal to four. So the final step is to substitute 𝑑 equals four into our model. We have then that 𝑃 of four is equal to 10000 over one plus nine 𝑒 to the negative four ln three. This actually works out very nicely, which you can see if you apply laws of logarithms in the denominator. We get 10000 over 10 over nine, which is 10000 times nine over 10, which is equal to 9000. So we find that the population after another three years, that is, the population four years after we’ve started, is 9000.

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