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The population P (in thousands) of a certain city from 2000 through 2014 can be modeled by P = 120.7ekt, where t represents the year, with t = 0 corresponding to 2000. In 2009, the population of the city was about 167,025.(a) Find the value of k. (Round your answer to four decimal places.)k=2)(b) Use the model to predict the populations of the city (in thousands) in 2020 and 2025. (Round your answers to three decimal places.)2020 P = thousand people2025 P = thousand people(c) According to the model, during what year will the population reach 220,000?anser=

Respuesta :

SOLUTION

Write out the given expression

[tex]\begin{gathered} P=120.7e^{kt} \\ t=\text{The year} \end{gathered}[/tex]

In 2009, the population is 167,025 implies

[tex]\begin{gathered} t=9 \\ P=167,025 \end{gathered}[/tex]

To obtain the value of K, substitute the given value into the expression

[tex]\begin{gathered} 167,025=120.7e^{k(9)} \\ 167,025=120.7e^{9k} \end{gathered}[/tex]

Divide both sides by 120.7

[tex]\begin{gathered} \frac{167025}{120.7}=e^{9k} \\ \text{Then} \\ 1383.8028=e^{9k} \\ \end{gathered}[/tex]

Then take natural logarithm of both sides

[tex]\begin{gathered} ln1383.8028=\ln e^{9k} \\ \text{Then} \\ ln1383.8028=9k \\ \end{gathered}[/tex]

Then, Divide both sides by 9

[tex]\begin{gathered} k=\frac{ln1383.8028}{9} \\ \text{Then} \\ k=0.8036 \end{gathered}[/tex]

Therefore

The value of k is 0.8036

B).Using the given model, the population in 2020 will be

[tex]\begin{gathered} \text{For the 2020, } \\ t=20 \\ k=0.8036 \end{gathered}[/tex]

The population will be

[tex]\begin{gathered} P=120.7e^{kt} \\ P=120.7e^{0.8036(20)} \\ P=1152625393 \end{gathered}[/tex]

Therefore, the population in 2020 is 1152625393 thousand people

The Population for 2025 will be

[tex]\begin{gathered} k=0.8036 \\ t=25 \end{gathered}[/tex]

Then

[tex]\begin{gathered} P=120,7e^{0.8036(25)} \\ P=120.7e^{20.09} \\ P=120.7\times530855280.2 \\ P=6.407\times10^{10} \end{gathered}[/tex]

Therefore, the population for 2025 is 6.407x10^10

C).P=220,000

[tex]\begin{gathered} P=120.7e^{^{kt}} \\ \text{Where } \\ P=220000 \\ k=0.8036 \\ t=\text{?} \end{gathered}[/tex]

Then substitute the values into the expression

[tex]\begin{gathered} 220000=120.7e^{0.8036t} \\ \text{Divide both sides by 120.7} \\ \frac{220000}{120.7}=\frac{120.7e^{0.8036t}}{120.7} \\ \\ 1822.7009=e^{0.8036t} \\ \end{gathered}[/tex]

Take the natural logarithm of the equation in the last line

[tex]\begin{gathered} \ln 1822.7009=\ln e^{0.8036t} \\ \ln 1822.7009=0.8036t \end{gathered}[/tex]

Then divide both sides by t

[tex]\begin{gathered} \frac{\ln 1822.7009}{0.8036}=\frac{0.8036t}{0.8036} \\ \frac{7.5081}{0.8036}=t \\ t=9.34 \end{gathered}[/tex]

Hence

The population will reach 220000 in 9years