Answer:
[tex]n = 500e^{-4.1589x}[/tex]
Step-by-step explanation:
Using the exponential growth equation [tex]N(t) = N_0e^{-kt}[/tex] where;
N₀ is the initial population of the bacteria
N(t) is the population of the bacteria during time t
If a biologist puts an initial population of 500 bacteria into a growth plate, this means that N₀ = 500 at t = 0
[tex]N(t) = N_0e^{-kt}\\N(t) = 500e^{-k(0)}\\N(t) = 500e^0\\N(t) = 500[/tex]
If the population is expected to double every 4 hours, this means when t = 4, N₀ = 2(500) = 1000
[tex]N(t) = N_0e^{-kt}\\500 = 1000e^{-4k}\\\frac{500}{1000} = e^{-4k}\\0.5 = e^{-4k}\\ln(0.5) = lne^{-4k}\\ln(0.5) = -4k\\k = \frac{ln0.5}{-4} \\k = 0.1733[/tex]
The equation that will give the expected number of bacteria, n, after x days, can be gotten by substituting N(t) = n and t = x days = 24x hours
[tex]N(t) = N_0e^{-kt}\\n = N_0e^{-0.1733(24x)}\\n = 500e^{-4.1589x}[/tex]
Hence the equation that will give the expected number of bacteria, n, after x days is [tex]n = 500e^{-4.1589x}[/tex]
