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Sagot :
a) Doubling time of the population.
Initial population = 4
Doubling = 2 x 4 = 8
[tex]8=4e^{0.03t}[/tex]Then, solve for t:
[tex]\begin{gathered} \frac{8}{4}=\frac{4e^{0.03t}}{4} \\ 2=e^{0.03t} \end{gathered}[/tex]Apply the exponent laws:
[tex]\begin{gathered} \ln 2=0.03t \\ \frac{\ln2}{0.03}=\frac{0.03t}{0.03} \\ t=23.1\approx23 \end{gathered}[/tex]Answer a: 23 years
b) Initial population = 4
Triple the population = 3 x 4 = 12
Therefore:
[tex]\begin{gathered} 12=4e^{0.03t} \\ \frac{12}{4}=\frac{4e^{0.03t}}{4} \\ 3=e^{0.03t} \\ \ln 3=0.03t \\ \frac{\ln 3}{0.03}=\frac{0.03t}{0.03} \\ t=36.6\approx37 \end{gathered}[/tex]Answer b: 37 years
c) Initial population = 4
Quadruple the population = 4 x 4 = 16
So:
[tex]\begin{gathered} 16=4e^{0.03t} \\ \frac{16}{4}=\frac{4e^{0.03t}}{4} \\ 4=e^{0.03t} \\ \ln 4=0.03t \\ \frac{\ln 4}{0.03}=\frac{0.03t}{0.03} \\ t=46.2\approx46 \end{gathered}[/tex]Answer c: 46 years
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