MATH SOLVE

3 months ago

Q:
# A. in 2000, the population of a country was approximately 5.755.75 million and by 20282028 it is projected to grow to 88 million. use the exponential growth model upper a equals upper a 0 e superscript kta=a0ekt, in which t is the number of years after 2000 and upper a 0a0 is in millions, to find an exponential growth function that models the data.b.b. by which year will the population be 77 million?

Accepted Solution

A:

A. in 2000, the population of a country was approximately 5.75 million and by 2028 it is projected to grow to 88 million. use the exponential growth model to find an exponential growth function that models the data.

B. By which year will the population be 77 million?

To solve this we are going to use the exponential growth model: [tex]A=A_{0}e^{kt}[/tex]

where

[tex]A[/tex] is the final population after [tex]t[/tex] years

[tex]A_{0}[/tex] is the initial population

[tex]e[/tex] is the Euler's constant

[tex]k[/tex] is the growth rate

[tex]t[/tex] is the time in years

A. We know for our problem that [tex]A=88[/tex], [tex]A_{0}=5.75[/tex], and [tex]t=2028-2000=28[/tex]. So lets replace those values in our exponential growth model to find [tex]k[/tex]:

[tex]A=A_{0}e^{kt}[/tex]

[tex]88=5.75e^{28k}[/tex]

[tex]e^{28k}= \frac{88}{5.75} [/tex]

[tex]ln(e^{28k})=ln( \frac{88}{5.75} )[/tex]

[tex]28k=ln( \frac{88}{5.75} )[/tex]

[tex]k= \frac{ln( \frac{88}{5.75}) }{28} [/tex]

[tex]k=0.09743[/tex]

Now that we have [tex]k[/tex], we can find an exponential growth function that models the data by replacing [tex]A_{0}[/tex] and [tex]k[/tex] in the exponential growth model:

[tex]A=5.75e^{0.09743t}[/tex]

We can conclude that the equation that models the data is: [tex]A=5.75e^{0.09743t}[/tex].

B. We know for a our problem that [tex]A=77[/tex] and [tex]A_{0}=5.75[/tex]. We also know from our previous calculation that [tex]k=0.09743[/tex]. So lets replace those values in our equation to find [tex]t[/tex]:

[tex]A=5.75e^{0.09743t}[/tex]

[tex]77=5.75e^{0.09743t}[/tex]

[tex]e^{0.09743t}= \frac{77}{5.75} [/tex]

[tex]ln(e^{0.09743t})=ln( \frac{77}{5.75})[/tex]

[tex]0.09743t=ln( \frac{77}{5.75})[/tex]

[tex]t= \frac{ln( \frac{77}{5.75}) }{0.09743} [/tex]

[tex]t=26.6[/tex]

We can conclude that the population of the country will reach 77 million by July of 2026.

B. By which year will the population be 77 million?

To solve this we are going to use the exponential growth model: [tex]A=A_{0}e^{kt}[/tex]

where

[tex]A[/tex] is the final population after [tex]t[/tex] years

[tex]A_{0}[/tex] is the initial population

[tex]e[/tex] is the Euler's constant

[tex]k[/tex] is the growth rate

[tex]t[/tex] is the time in years

A. We know for our problem that [tex]A=88[/tex], [tex]A_{0}=5.75[/tex], and [tex]t=2028-2000=28[/tex]. So lets replace those values in our exponential growth model to find [tex]k[/tex]:

[tex]A=A_{0}e^{kt}[/tex]

[tex]88=5.75e^{28k}[/tex]

[tex]e^{28k}= \frac{88}{5.75} [/tex]

[tex]ln(e^{28k})=ln( \frac{88}{5.75} )[/tex]

[tex]28k=ln( \frac{88}{5.75} )[/tex]

[tex]k= \frac{ln( \frac{88}{5.75}) }{28} [/tex]

[tex]k=0.09743[/tex]

Now that we have [tex]k[/tex], we can find an exponential growth function that models the data by replacing [tex]A_{0}[/tex] and [tex]k[/tex] in the exponential growth model:

[tex]A=5.75e^{0.09743t}[/tex]

We can conclude that the equation that models the data is: [tex]A=5.75e^{0.09743t}[/tex].

B. We know for a our problem that [tex]A=77[/tex] and [tex]A_{0}=5.75[/tex]. We also know from our previous calculation that [tex]k=0.09743[/tex]. So lets replace those values in our equation to find [tex]t[/tex]:

[tex]A=5.75e^{0.09743t}[/tex]

[tex]77=5.75e^{0.09743t}[/tex]

[tex]e^{0.09743t}= \frac{77}{5.75} [/tex]

[tex]ln(e^{0.09743t})=ln( \frac{77}{5.75})[/tex]

[tex]0.09743t=ln( \frac{77}{5.75})[/tex]

[tex]t= \frac{ln( \frac{77}{5.75}) }{0.09743} [/tex]

[tex]t=26.6[/tex]

We can conclude that the population of the country will reach 77 million by July of 2026.