In this problem, we have an exponential function of the form
y=a(b)^x
where
y is the population of San Diego
x number of years since 2012
so
Part a
2012 represents x=0, y=1,309,000
a=1,309,000
For the year 2016 ------> x=2016-2012=4 years
y=1,375,000
therefore
![1,375,000=1,309,000(b)^4](https://img.qammunity.org/2023/formulas/mathematics/college/cb3rzzazboq7ngeog7dw8hd2qa7l7k1raj.png)
solve for b
b=1.0124
b=1+r
substitute and solve for r
r=1.0124-1
r=0.0124
percentage
r=1.24%
Part b
2016-2020
in this part
a=1,375,000
x is the number of years since 2016
For the year 2020 -----> x=2020-2016=4 years
y=1,386,000
substitute
![1,386,000=1,375,000(b)^4](https://img.qammunity.org/2023/formulas/mathematics/college/y9m8anflsqhczevvcn1d59q5p756sdxotg.png)
solve for b
b=1.0020
b=1+r
r=1.0020-1
r=0.0020
percentage
r=0.20%
Part c
Compare the growth rate each period
2012-2016 ------> r=1.24%
2016-2020 -----> r=0.20%
that means ----> in the second period the growth rate has decreased than the first period
Part d
the current growth rate is r=0.20%
so
the equation is
![y=1,375,000(1.002)^x](https://img.qammunity.org/2023/formulas/mathematics/college/vnm98qje9orko53hdzja9a35nqg7vof087.png)
where
x is the number of years since 2016
so
For y=1,500,000
substitute
![1,500,000=1,375,000(1.002)^x](https://img.qammunity.org/2023/formulas/mathematics/college/2fbbeaqt5vkuh8kne9nrobcw6y9ygv8113.png)
solve for x
![(1,500,000)/(1,375,000)=(1.002)^x](https://img.qammunity.org/2023/formulas/mathematics/college/fm0fp1jnmli36y90gobkh196okg2toi903.png)
apply log on both sides
![\log ((1,500,000)/(1,375,000))=x\cdot\log (1.002)](https://img.qammunity.org/2023/formulas/mathematics/college/atlanxhexqxu4s257j6c4va5xxa397j25g.png)
x=43.5 years
therefore
the year is
2016+43.5=2059.5 -------> approximate the year 2060