A town has a population of 8000 and grows at 3% every year. To the nearest tenth of a year, how long will it be until the population will reach 16300?

Answer 1

16300 = 8000e^.03t

163/80 = e^.03t

ln(163/80) = .03t

t = ln(163/80)/.03

t = about 23.724 years, whether it's a human or rabbit population

t = 23 years, 8 months, and almost 21 days

population growth tends to be continuous, similar to continuoulsy compounded interest

but if you interpreted the problem to mean 3% compounded only annually, then

163/80 =(1+.03)^t

163/80 =(1.03)^t

l2.0375 = 1.03^t

t = 24.1 years

but populations reallistically don't grow that way

Answer 2

We can use the following exponential equation to model this problem:

Initial Population * Rate of Annual Growth ^ Years = Final Population

8000 * (1.03)^t = 16300, where t is the number of years that have passed

Note that: 1.03^t = 16300/8000 = 2.0375

We can apply the natural log function here, because we know that if a = b, then ln(a) = ln(b), so:

ln(1.03^t) = ln(2.0375)

By natural log rules, we know that ln(1.03^t) = t * ln(1.03)

t * ln(1.03) = ln(2.0375)

t = ln(2.0375)/ln(1.03), which means t is approximately equal to 24.1 years.