Question

A country's population in 1994 was 182 million. In 2002 it was 186 million. Estimate the population in 2004 using the exponential growth formula. Round your answer to the nearest million. P=ae^kt

Answer 1

$\bf =ae^{kt}\qquad \begin{cases} 1994\impliedby \textit{year 0, starting point}\\ t=0\qquad P=182 \end{cases}\implies 182=ae^{k0} \\\\\\ 182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a \\\\\\ thus\qquad P=182e^{kt}\\\\ -------------------------------$

$\bf P=182e^{kt}\qquad \begin{cases} 2002\impliedby \textit{8 years later}\\ t=8\qquad P=186 \end{cases}\implies 186=182e^{k8} \\\\\\ \cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k \\\\\\ \cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}$

what's the population in 2004?  well,  from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004