Question

At the start of a research study, a colony of penguins had a population of 20,000. One year later, it had a population of 21,200.Assuming the population of the colony has grown exponentially, which expression best models thepopulation? Let t represent the time in years from the start of the research study.1,200(1.015)^t20,000 (1.06)^4t21,200 (1.012)^t20,000 (1.06)^tAssuming the colony continues to grow at the same rate, what will the population of the colony be 4 years after the start of the research study?Round your answer to the nearest whole number.

Answer 1

An exponential function is generally expressed as

`\begin{gathered} y=a(b)^t\text{ ----- equation 1} \\ \end{gathered}`

Given that in a research study, a colony of penguins had a population of 20,000.

This implies that

`\begin{gathered} when\text{ t=0,} \\ 20,000=ab^0 \\ \Rightarrow20000=a*1\text{ \lparen where b}^0=1) \\ thus, \\ a=20000 \end{gathered}`

Substitute the value of a into equation 1.

Thus,

`y=20000(b)^t\text{ ----- equation 2}`

One year later, it had a population of 21,200. This implies that when t equals 1, we substitute the values of 21200 and 1 for y and t respectively into equation 2.

This gives

`\begin{gathered} 21200=20000(b)^1 \\ \Rightarrow21200=20000b \\ divide\text{ both sides by the coefficient of b, which is b.} \\ thus, \\ (21200)/(20000)=(20000b)/(20000) \\ \Rightarrow b=1.06 \end{gathered}`

Substitute the obtained value of b into equation 2.

Thus, the expression that best models the population is

`20,000(1.06)^t`

Assuming the colony grows at the same rate, the population of the colony after 4 years is evaluated by solving for y when the value of t is 4.

Thus,

`\begin{gathered} y=20,000(1.06)^t \\ when\text{ t=4, we have} \\ y=20,000(1.06)^4 \\ =20000*(1.06)^4 \\ =20000*1.26247696 \\ \Rightarrow y=25249.5392 \\ \therefore y=25250\text{ \lparen nearest whole number\rparen} \end{gathered}`

Hence, after 4 years the population of the colony will be 25250 penguins (nearest whole number).