# Exponential equations to model population growth

## Exponential growth is modeled an exponential equation

The population of a species that grows exponentially over time can be modeled by

???P(t)=P_0e^{kt}???

where ???P(t)??? is the population after time ???t???, ???P_0??? is the original population when ???t=0???, and ???k??? is the growth constant.

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This type of growth is usually found in smaller populations that aren’t yet limited by their environment or the resources around them. In a small population, growth is nearly constant, and we can use the equation above to model population.

When a population becomes larger, it’ll start to approach its carrying capacity, which is the largest population that can be sustained by the surrounding environment. At that point, the population growth will start to level off. If the population ever exceeds its carrying capacity, then growth will be negative until the population shrinks back to carrying capacity or lower. To model population growth and account for carrying capacity and its effect on population, we have to use the equation

???\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)???

where ???M??? is the carrying capacity of the population.

Let’s try an example with a small population that has normal growth.

## How long does it take for a population to double?

Example

A bacteria population increases tenfold in ???8??? hours. Assuming normal growth, how long did it take for the population to double?

If we use hours as the units for ???t???, then we can say right away that

???P(t)=P_0e^{k(8)}???

We weren’t given initial population ???P_0???, and we’ve been asked to find ???P(t)???, so we can’t plug in for either of those variables. Fortunately, we’ve been told that the population grows to ???10??? times its original size in ???8??? hours. If we say that ???P_0??? is the original population, and ???10P_0??? is ???10??? times the original population, then

???10P_0=P_0e^{k(8)}???

???\frac{10P_0}{P_0}=e^{k(8)}???

???10=e^{8k}???

???\ln{10}=\ln{e^{8k}}???

???\ln{10}=8k???

???k=\frac{\ln{10}}{8}???

Now that we have a value for ???k???, we can can figure out how long it took for the population to double. If we say that ???P_0??? is the original population, and ???2P_0??? is double the original population, then

???2P_0=P_0e^{\frac{\ln{10}}{8}t}???

???\frac{2P_0}{P_0}=e^{\frac{\ln{10}}{8}t}???

???2=e^{\frac{\ln{10}}{8}t}???

???\ln{2}=\ln{e^{\frac{\ln{10}}{8}t}}???

???\ln{2}=\frac{\ln{10}}{8}t???

???t=\frac{8\ln{2}}{\ln{10}}???

???t\approx2.41???

The bacteria’s population reached double its original size in about ???2.41??? hours.

Now we’ll do an example with a larger population, in which carrying capacity is effecting its growth rate.

When a population becomes larger, it’ll start to approach its carrying capacity, which is the largest population that can be sustained by the surrounding environment.

Example

Write a logistic growth equation and find the population after ???5??? years for a group of ducks with an initial population of ???P=1,500???, and a carrying capacity of ???M=16,000???. The duck population after ???2??? years is ???2,000???.

We’ll start by plugging what we know into the logistic growth equation. With ???P=1,500??? and ???M=16,000???, we get

???\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)???

???\frac{dP}{dt}=k(1,500)\left(1-\frac{1,500}{16,000}\right)???

???\frac{dP}{dt}=1,500k\left(1-\frac{3}{32}\right)???

???\frac{dP}{dt}=1,500k\left(\frac{29}{32}\right)???

???\frac{dP}{dt}=\frac{10,875k}{8}???

This is the logistic growth equation. Now we need to find population after ???5??? years. We’ll treat this like a separable differential equations problem, integrate both sides, and solve for ???P??? as a function of time ???t???.

???dP=\frac{10,875k}{8}\ dt???

???\int\ dP=\int\frac{10,875k}{8}\ dt???

???P+C_1=\frac{10,875kt}{8}+C_2???

???P=\frac{10,875kt}{8}+C_2-C_1???

???P=\frac{10,875kt}{8}+C???

We have the initial condition ???P(0)=1,500???, so

???1,500=\frac{10,875k(0)}{8}+C???

???1,500=0+C???

???C=1,500???

Therefore,

???P=\frac{10,875kt}{8}+1,500???

We were also told in the problem that the duck population after ???2??? years is ???2,000???. Plugging in this information, we get

???2,000=\frac{10,875k(2)}{8}+1,500???

???500=\frac{10,875k}{4}???

???2,000=10,875k???

???k=\frac{2,000}{10,875}???

???k=\frac{16}{87}???

Since we want to find the duck population after ???5??? years, we’ll plug in the value we just found for ???k???, plus ???t=5???.

???P=\frac{10,875\left(\frac{16}{87}\right)(5)}{8}+1,500???

???P=\frac{\frac{870,000}{87}}{8}+1,500???

???P=\frac{10,000}{8}+1,500???

???P=1,250+1,500???

???P=2,750???

The duck population reached ???2,750??? after ???5??? years.