# Question #83d0e

Dec 7, 2014

$A \left(t\right) = 965.1 {e}^{0.016 t}$

Since the population is 753 million people, we have

$965.1 {e}^{0.016 t} = 753$

by dividing by $965.1$,

$\implies {e}^{0.016 t} = \frac{753}{965.1}$

by taking the natural log,

$\implies 0.016 t = \ln \left(\frac{753}{965.1}\right)$

by dividing by $0.016$,

$\implies t = \frac{\ln \left(\frac{753}{965.1}\right)}{0.016} \approx - 15.5$

Hence, it was $15.5$ years before 2003.

I hope that this was helpful.