Answer:
the time at which P =15 is [tex]{t_o = 0.54161 \ years}[/tex]
the time at which P =20 is [tex]{t_1 = 0.59914 \ years}[/tex]
the doubling time [tex]{t = 0.138629 \ years}[/tex]
Step-by-step explanation:
From the information we are being provided with:
The quantity P i.e [tex]P(t) = e ^{5t}[/tex]
In order to determine the time at which P = 15, we have the following:
[tex]t= t_o[/tex] when P = 15
∴ [tex]e^{5t_o} = 15[/tex]
[tex]5t_o= In (15)[/tex]
[tex]t_o = \dfrac{1}{5}In 15[/tex]
[tex]t_o = \dfrac{1}{5} \times 2.70805[/tex]
[tex]{t_o = 0.54161 \ years}[/tex]
Hence, the time at which P =15 is [tex]{t_o = 0.54161 \ years}[/tex]
In order to determine the time at which P = 20, we have the following:
[tex]t= t_1[/tex] when P = 20
∴ [tex]e^{5t_1} = 20[/tex]
[tex]5t_1= In (20)[/tex]
[tex]t_1= \dfrac{1}{5}In (20)[/tex]
[tex]t_1 = \dfrac{1}{5} \times 2.9957[/tex]
[tex]{t_1 = 0.59914 \ years}[/tex]
Hence, the time at which P =20 is [tex]{t_1 = 0.59914 \ years}[/tex]
The doubling time at t = 0, mean P = 2
∴ [tex]e^{5t} = 2[/tex]
[tex]5t= In (2)[/tex]
[tex]t= \dfrac{1}{5}In(2)[/tex]
[tex]t = \dfrac{1}{5} \times0.693147[/tex]
[tex]{t = 0.138629 \ years}[/tex]
Hence, the doubling time [tex]{t = 0.138629 \ years}[/tex]
