Answer :
Answer:
[tex] f(t) = 6000 (\frac{1}{2})^t^/^3 [/tex]
Step-by-step explanation:
Let's take the equation:
[tex] y = y_0 e^-^k^t[/tex]
Given the initial population of bacteria = 6000,
[tex] y = 6000e^-^k^t[/tex]
Population of bacteria after 3 hours is 3000.
y(3) = 3000
Thus,
[tex] 6000e^-^k^3 = 3000[/tex]
[tex] e^-^k^3 = \frac{3000}{6000}[/tex]
[tex] e^-^k^3 = \frac{1}{2}[/tex]
[tex] (e^-^k)^3 = \frac{1}{2}[/tex]
[tex] e^-^k= (\frac{1}{2})^1^/^3[/tex]
Let's substitute [tex] (\frac{1}{2})^1^/^3 [/tex] for [tex] e^-^k[/tex] in [tex] y = 6000e^-^k^t [/tex]
Therefore, we have:
[tex] 6000e^-^k^t [/tex]
[tex] = 6000 [(\frac{1}{2}) ^1^/^3]^t[/tex]
[tex] = 6000 [\frac{1}{2}] ^t^/^3 [/tex]