Answer:
The population in 20 years will be 781.
Explanation:
The population of the town can be modeled by the following equation:

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate, as a decimal.
The initial population of 500 increases by 25% in 10 years.
This means that

We use this to find r.



Applying ln to both sides




So

What will be the population in 20 years
This is P(20)

The population in 20 years will be 781.