Question

A population doubles every 27 years. Assuming exponential growth find the following:(a) The annual growth rate:___% help (numbers)(b) The continuous growth rate is ___% per year help (numbers)

Answer 1

a) Exponential growth formula

$y=a(1+r)^t$

where:

• y: final amount

,

• a: initial amount

,

• r: annual growth rate, as a decimal

,

• t: time in years

Given that the population doubles every 27 years, then y = 2a (the final amount is two times the initial amount), when t = 27 years. Substituting this into the formula and solving for r:

$\begin{gathered} 2a=a(1+r)^{27} \\ \frac{2a}{a}=\frac{a(1+r)^{27}}{a} \\ 2=(1+r)^{27} \\ \log _{10}2=\log _{10}\lbrack(1+r)^{27}\rbrack \\ \log _{10}2=27\log _{10}(1+r) \\ \frac{\log _{10}2}{27}=\frac{27\log _{10}(1+r)}{27} \\ \frac{\log_{10}2}{27}=\log _{10}(1+r) \\ 0.0111\approx\log _{10}(1+r) \\ 10^{0.0111}\approx1+r \\ 10^{0.0111}-1\approx1+r-1 \\ 0.026\approx r \end{gathered}$

Expressed as a percentage, the value of r is:

$r=0.026=0.026\cdot100=2.6\%$

The annual growth rate: 2.6%

b) Exponential growth can be also modeled by the next formula:

$P=P_0e^{kt}$

where:

• P: final population

,

• P₀: initial population

,

• k: continuous growth rate, as a decimal

,

• t: time in years

Similarly as before, substituting with P = 2P₀ and t = 27 years, and solving for k, we get:

$\begin{gathered} 2P_0=P_0e^{k\cdot27} \\ \frac{2P_0}{P_0}=e^{27k} \\ 2=e^{27k} \\ \ln 2=27k\ln e \\ \frac{\ln2}{27}=k \\ 0.0256\approx k \end{gathered}$

Expressed as a percentage, the value of k is:

$k=0.0256=0.0256\cdot100=2.56\%$

The continuous growth rate is 2.56%