Answer:
In the year 2010
Step-by-step explanation:
we have
[tex]A=200e^{0.033t}[/tex]
where
A is the population in thousands
t is the number of years since 1998
so
For A=297 thousands
substitute and solve for t
[tex]297=200e^{0.033t}[/tex]
simplify
[tex]1.485=e^{0.033t}[/tex]
Apply ln both sides
[tex]ln(1.485)=ln[e^{0.033t}][/tex]
[tex]ln(1.485)=(0.033t)ln(e)[/tex]
Remember that
[tex]ln(e)=1[/tex]
[tex]ln(1.485)=(0.033t)[/tex]
[tex]t=\frac{ln(1.485)}{0.033}[/tex]
[tex]t=12\ years[/tex]
1998+12=2010
