a. The continuous growth rate of the bacteria is 21%
b. The initial population of bacteria is 715
c. The culture will contain 2043 bacteria after 6 × 10⁻⁴ years
a. What is the continuous rate of growth of this bacteria population?
Since [tex]n(t) = 715e^{0.21t}[/tex] represents the number of bacteria in the culture.
This function is similar to an exponential function of the form [tex]y(t) = Ae^{\lambda t}[/tex] where λ = growth rate
Comparing n(t) and y(t), we see that λ = 0.21
So, the continuous growth rate of the bacteria is 0.21 = 0.21 × 100 %
= 21%
So, the continuous growth rate of the bacteria is 21%
b. What is the initial population of the culture?
Since [tex]n(t) = 715e^{0.21t}[/tex] represents the number of bacteria in the culture, the initial population of bacteria is obtained when t = 0.
So, [tex]n(t) = 715e^{0.21t}[/tex]
[tex]n(0) = 715e^{0.21(0)} \\= 715e^{0} \\= 715 X 1\\= 715[/tex]
So, the initial population of bacteria is 715
c. When will the culture contain 2043 bacteria?
To find the time when the number of bacteria will be 2043, this means n(t) = 2043.
Since [tex]n(t) = 715e^{0.21t}[/tex]
Making t subject of the formula, we have
t = ㏑[n(t)/715]/0.21
So, substituting n(t) = 2043 into the equation, we have
t = ㏑[n(t)/715]/0.21
t = ㏑[2043/715]/0.21
t = ㏑[2.8573]/0.21
t = 1.05/0.21
t = 4.99
t ≅ 5 hours
Converting this to years, we have t = 5 h × 1 day/24h × 1 year/365 days
= 5/8760
= 5.7 × 10⁻⁴ years
≅ 6 × 10⁻⁴ years
So, the culture will contain 2043 bacteria after 6 × 10⁻⁴ years
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