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( (1200)/(1110.9) \right)=ln(e^(2k))\implies ln\left( (1200)/(1110.9) \right)=2k \\\\\\ \cfrac{ln\left( (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}`