Question

Need to know the population in 2015

Answer 1

$\bf P=1110.9e^{kt}\quad \begin{cases} P=1200 &\textit{in thousands}\\ t=2 &\textit{in 2002} \end{cases}\implies 1200=1110.9e^{k2} \\\\\\ \cfrac{1200}{1110.9}=e^{2k}\implies ln\left( \frac{1200}{1110.9} \right)=ln(e^{2k})\implies ln\left( \frac{1200}{1110.9} \right)=2k \\\\\\ \cfrac{ln\left( \frac{1200}{1110.9} \right)}{2}=k\implies 0.0385755\approx k\implies 0.0386\approx k\\\\ -------------------------------$

$\bf P=1110.9e^{0.0386t}\qquad \begin{cases} t=14\\ \textit{year 2015} \end{cases}\implies P=1110.9e^{0.0386\cdot 14} \\\\\\ P\approx 1907.0747\implies about\ 1,907,075\textit{ once rounded up}$