Question

A species of frog’s population grows 24% every year. Suppose 100 frogs are released into a pond.10. Construct an exponential model for this population.11. How many frogs will there be in 5 years? (round to the nearest frog)12. How many frogs will there be in 10 years? (round to the nearest frog)13. How many years will it take for there to be at least 1000 frogs?

Answer 1

For this problem, we are informed that a population of frogs grows 24% every year, and we are asked to create a model for the growth of 100 frogs released into a pond.

The initial value of the population is equal to 100 frogs, so we have:

`P(0)=100`

After the first year, the population grew by 24%, so we have:

`\begin{gathered} P(1)=100\cdot(1+_{}(24)/(100)) \\ P(1)=P(0)\cdot1.24 \end{gathered}`

After the second year, the population grew again by 24%, but this time in relation to the population after the first year, so we have:

`\begin{gathered} P(2)=P(1)\cdot1.24 \\ P(2)=P(0)\cdot1.24\cdot1.24 \\ P(2)=P(0)\cdot1.24^2 \end{gathered}`

In the third year, the population increased again by the same rate, we can represent it as shown below:

`\begin{gathered} P(3)=P(0)\cdot1.24\cdot1.24^2 \\ P(3)=P(0)\cdot1.24^3 \end{gathered}`

We can now see a trend, after a certain year, the size of the population can be found by multiplying the initial population with the increase rate powered by the number of years that have passed. A general model would be:

`P(n)=P(0)\cdot1.24^n`

Since we know the value of the initial population, the complete model is:

`P(n)=100\cdot1.24^n`

Now we need to find the population after 5 years:

`\begin{gathered} P(5)=100\cdot1.24^5 \\ P(5)=100\cdot2.9316 \\ P(5)=293.16 \end{gathered}`

The population of frogs will be equal to 293 frogs after 5 years.

Now we need to find the population after 10 years:

`\begin{gathered} P(10)=100\cdot1.24^(10) \\ P(10)=100\cdot8.5944 \\ P(10)=859.44 \end{gathered}`

After 10 years, the population will be 859 frogs.

Now we need to determine how many years it will take to have at least 1000 frogs. For this, we need to replace P(n) with 1000, and solve for n.

`\begin{gathered} 1000=100\cdot1.24^n \\ 1.24^n=(1000)/(100) \\ 1.24^n=10 \\ \ln 1.24^n=\ln 10 \\ n\cdot\ln 1.24=\ln 10 \\ n=(\ln 10)/(\ln 1.24)=10.702 \end{gathered}`

The population will be at least 1000 frogs for any time greater than or equal to 10.702 years.