Question

10. A city has a population of 125,500 in the year 1989. In the year 2007, its population is 109, 185. A. Find the continuous growth/decay rate for this city. Be sure to show all your work.B. If the growth/decay rate continues, find the population of the city in the year 2021.C. In what year will the population of the city reach 97,890? Be sure to show all your work.

Answer 1

A.

To solve this question, we will use the compound interest formula.

Which is:

`\begin{gathered} A=P(1-(r)/(100))^(nt) \\ Since\text{ we are dealing with a yearly statistics, n = 1} \end{gathered}`
`\begin{gathered} \text{From 1989 to 2007, there is a year difference of 18 years} \\ t=18 \\ A=109,185 \\ P=125,500 \\ We\text{ are looking for the continuous growth rate (r)} \\ \text{Now, we will substitute all these given parameters into the formula } \\ \text{above.} \end{gathered}`
`\begin{gathered} 109,185=\text{ 125,500(1-}(r)/(100))^(18) \\ (195185)/(125500)=(125500)/(125500)(1-(r)/(100))^(18) \\ 0.87=(1-(r)/(100))^(18) \\ \text{take the natural logarithm of both sides:} \\ \ln 0.87=18\ln (1-(r)/(100)) \\ -0.1393=18\ln (1-(r)/(100)) \\ (-0.1393)/(18)=\ln (1-(r)/(100))_{}_{}_{}_{}_{} \\ -0.007737=\ln (1-(r)/(100)) \\ \end{gathered}`
`\begin{gathered} e^(-0.007737)=(1-(r)/(100)) \\ 0.9922=1-(r)/(100) \\ (r)/(100)=1-0.9922 \\ (r)/(100)=0.007707 \\ r=100*0.007707 \\ r=0.771\text{ \%} \end{gathered}`

The continuous decay rate is 0.771%

B.

Using the same formula:

`\begin{gathered} A=P(1-(r)/(100))^(nt) \\ t=2021-2007=14 \\ P=109,185 \\ n=1 \\ A=\text{?} \\ r=0.771 \\ \text{Substitute all the parameters into the formula above:} \end{gathered}`
`\begin{gathered} A=P(1-(r)/(100))^(nt) \\ A=109,185(1-(0.771)/(100))^(1*14) \\ A=109,185*0.89730607 \\ A=97,972.36 \\ A=97,972\text{ (to the nearest person)} \end{gathered}`

The population of the city in the year 2021 is 97,972.

C.

We will use the same formula:

`\begin{gathered} A=P(1-(r)/(100))^(nt) \\ A=97,890 \\ P=125,500 \\ r=0.771 \\ t=\text{?} \\ \text{Substitute all these parameters into the formula above:} \\ \end{gathered}`
`\begin{gathered} 97890=125,500(1-(0.771)/(100))^t^{} \\ (97890)/(125500)=(125500)/(125500)(0.99229)^t \\ 0.78=0.99229^t \\ \ln 0.78=t\ln 0.99229 \\ -(0.2485)/(\ln 0.99229)=t \\ t=32.101 \\ SO\text{ the year that the population will reach 97,890 will be:} \\ 1989+32.101=2021.101 \\ \text{Which is approximately year 2021.} \end{gathered}`