Question

A country's population in 1993 was 94 million. In 1999 it was 99 million. Estimate the population growth formula. Round your answer to the nearest million.
P=Ae^kt

Answer 1

I would let t=0 (the "start time" ) correspond to 1993
population at the start i.e. at t=0 (1993) is 94
At t=0, we can write
\[ A e^{k\cdot 0} = 94 \\ A e^0 = 94\]
remember anything to the zero power is 1. that is e^0 =1, so
\[ A e^0 = 94 \\ A= 94\]
in 1999, which is 6 years past 1993, t=6
we know at t=6 (i.e. 1999), P = 99

write
\[ 94 e^{6k} = 99\]

Answer 2

we now can solve for k. first divide both sides by 94
\[ 94 e^{6k} = 99 \\ e^{6k} = \frac{99}{94} \]
"take the natural log" of both sides
\[ \ln\left(e^{6k} \right) = \ln\left( \frac{99}{94} \right) \\
6k = \ln\left( \frac{99}{94} \right) \\
k = \frac{1}{6}\ \ln\left( \frac{99}{94} \right)
\]
we will need to use a calculator to find a decimal value for k

Answer 3

we will find k= 0.00863751131

\[ P= 94 e^{0.00864 t} \]