Question

The population of a city is P(t) = 4e^0.03t(in millions), where t is measured in years.(a) Calculate the doubling time of the population.(b) How long does it take for the population to triple in size?(c) How long does it take for the population to quadruple in size?

Answer 1

a) Doubling time of the population.

Initial population = 4

Doubling = 2 x 4 = 8

`8=4e^(0.03t)`

Then, solve for t:

`\begin{gathered} (8)/(4)=(4e^(0.03t))/(4) \\ 2=e^(0.03t) \end{gathered}`

Apply the exponent laws:

`\begin{gathered} \ln 2=0.03t \\ (\ln2)/(0.03)=(0.03t)/(0.03) \\ t=23.1\approx23 \end{gathered}`

Answer a: 23 years

b) Initial population = 4

Triple the population = 3 x 4 = 12

Therefore:

`\begin{gathered} 12=4e^(0.03t) \\ (12)/(4)=(4e^(0.03t))/(4) \\ 3=e^(0.03t) \\ \ln 3=0.03t \\ (\ln 3)/(0.03)=(0.03t)/(0.03) \\ t=36.6\approx37 \end{gathered}`

Answer b: 37 years

c) Initial population = 4

Quadruple the population = 4 x 4 = 16

So:

`\begin{gathered} 16=4e^(0.03t) \\ (16)/(4)=(4e^(0.03t))/(4) \\ 4=e^(0.03t) \\ \ln 4=0.03t \\ (\ln 4)/(0.03)=(0.03t)/(0.03) \\ t=46.2\approx46 \end{gathered}`

Answer c: 46 years