Answer:
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
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}) = 0.0287t\\\\t = \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}) = 0.0287t\\\\t = 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