The given equation is

Where A is the population from 2000 in t years
since the population in t years is 28.3 million, then
Substitute A by 28.3 to find t

Divide both sides by 23.1

Insert ln for both sides

Use the rule


Divide both sides by 0.0152 to find t

Round it to the nearest year, then
t = 14
Add 14 to 2000 to find the year
2000 + 14 = 2014, then
The population was 28.3 million at 2014