(a) We have the equation \(\displaystyle{P}={C}{e}^{{k}}{t}\) where t is the number of years since 1960. Setting t= 0, we have 181 = Ce^0 = C. To find the rate k, we can tse
the 1970 data:
\(\displaystyle{205}={181}{e}^{{10}}{k}\)

\(\displaystyle{k}=\frac{{1}}{{10}}{\left({\ln{{\left(\frac{{205}}{{181}}\right)}}}\right)}={0.1245}\)

This gives us the model \(\displaystyle{P}{1}={181}{e}^{{0.01245}}{t}\)

(b) Using a graphing utiliti gives us the model \(\displaystyle{P}{2}={182.32}{e}^{{0.0109}}{t}\)

(c) P1 i black, while P2 is graphed in blue:

(d) Setting P2=320, we can then solve for t to get:

\(\displaystyle{t}=\frac{{\ln{{1.7552}}}}{{0.0109}}\sim{51}\) years, or in the year 2011.

\(\displaystyle{k}=\frac{{1}}{{10}}{\left({\ln{{\left(\frac{{205}}{{181}}\right)}}}\right)}={0.1245}\)

This gives us the model \(\displaystyle{P}{1}={181}{e}^{{0.01245}}{t}\)

(b) Using a graphing utiliti gives us the model \(\displaystyle{P}{2}={182.32}{e}^{{0.0109}}{t}\)

(c) P1 i black, while P2 is graphed in blue:

(d) Setting P2=320, we can then solve for t to get:

\(\displaystyle{t}=\frac{{\ln{{1.7552}}}}{{0.0109}}\sim{51}\) years, or in the year 2011.