In 2012, the population of a city was 5.82 million. The exponential growth rate was 2.95% per year. a) Find the exponential growth function.
b) Estimate the population of the city in 2018.
c) When will the population of the city be million?
d) Find the doubling time.

a) P(t) = 6.57e ^{0.0287t}P(t)=6.57e

0.0287t

b) P(6) = 7,805\ millionP(6)=7,805 million

c) t = 14.64\ yearst=14.64 years

d) t = 24.15\ yearst=24.15 years

Step-by-step explanation:

a) The function of exponential growth for a population has the following formula:

P(t) = p_0e ^{rt}P(t)=p

0

e

rt

In this equation:

p_0p

0

is the initial population

r is the growth rate

t is the time in years

In this problem we know that

r = 2.87\% = 0.0287r=2.87%=0.0287

p_0= 6.57p

0

=6.57 million in the year 2012.

So the equation is:

P(t) = 6.57e ^{0.0287t}P(t)=6.57e

0.0287t

Where t = 0t=0 represents the year 2012

b) If t = 0t=0 in 2012, then in 2018 t = 6t=6

The population in 2018 is:

\begin{gathered}P(t = 6) = 6.57e ^{0.0287(6)}P(6) = 7,805\ million\end{gathered}

P(t=6)=6.57e

0.0287(6)

P(6)=7,805 million

c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.

\begin{gathered}P(t) = 10 = 6.57e ^{0.0287t}frac{10}{6.57} = e ^{0.0287t}ln(\frac{10}{6.57}) = 0.0287tt = \frac{ln(\frac{10}{6.57})}{0.0287}\end{gathered}

P(t)=10=6.57e

0.0287t

6.57

10

=e

0.0287t

ln(

6.57

10

)=0.0287t

t=

0.0287

ln(

6.57

10

)

t = 14.64\ yearst=14.64 years

d) The function is doubled when P(t) = 2p_0P(t)=2p

0

P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}P(t)=2(6.57)=13.14=6.57e

0.0287t

We solve for t.

\begin{gathered}\frac{13.14}{6.57} = e ^{0.0287t}ln(\frac{13.14}{6.57}) = 0.0287tt = ln{13.14}{6.57})}{0.0287})\

end

{gathered}

6.57

13.14

=e

0.0287t

ln(

6.57

13.14

)=0.0287t

t=

0.0287

ln(

6.57

13.14

)

)

t = 24.15\ yearst=24.15 years

10 down (bang1)

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hoped i lol