The given equation is
[tex]A=23.1e^{0.0152t}[/tex]
Where A is the population from 2000 in t years
since the population in t years is 28.3 million, then
Substitute A by 28.3 to find t
[tex]28.3=23.1e^{0.0152t}[/tex]
Divide both sides by 23.1
[tex]\begin{gathered} \frac{28.3}{23.1}=\frac{23.1}{23.1}e^{0.0152t} \\ \frac{283}{231}=e^{0.0152t} \end{gathered}[/tex]
Insert ln for both sides
[tex]\ln (\frac{283}{231})=\ln (e^{0.0152t})[/tex]
Use the rule
[tex]\ln (e^n)=n[/tex][tex]\ln (\frac{283}{231})=0.0152t[/tex]
Divide both sides by 0.0152 to find t
[tex]\begin{gathered} \frac{\ln (\frac{283}{231})}{0.0152}=\frac{0.0152t}{0.0152} \\ 13.35718336=t \end{gathered}[/tex]
Round it to the nearest year, then
t = 14
Add 14 to 2000 to find the year
2000 + 14 = 2014, then
The population was 28.3 million at 2014
