Question

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth or decay. b) Find the number of bacteria after 2 days c) How many hours will it take for the number of bacteria to reach 4000? d) How long will it take for the number of bacteria to triple?

Answer 1

a)
`N_0=250\; k=0.15`

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies
`t=0`. And
`N(0)=N_0`. The process is:

`N(t)=250 e^(0.15t)\\N(0)=250 e^(0.15(0))\\ N_0=250e^(0)\\N_0=250\cdot1\\ N_0=250`

b) After 2 days means
`t=48`. So, we just replace and operate:

`N(t)=250 e^(0.15t)\\N(48)=250 e^(0.15(48))\\ N(48)=250e^(7.2)\\N(48)=334,858\;\text{bacteria}`

c)
`N(t_1)=4000; \;t_1=?`

`N(t)=250 e^(0.15t)\\4000=250 e^(0.15t_1)\\ (4000)/(250)= e^(0.15t_1)\\16= e^(0.15t_1)\\ ln(16)= \ln{e^(0.15t_1)} \\  ln(16)=0.15t_1 \\ (ln(16))/(0.15)=t_1=4.67\approx 5\;h`

d)
`t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)`

`N(t)=250 e^(0.15t)\\ 750=250 e^(0.15t_2) \\ ln(3) =\ln{e^(0.15t_2)}\\ t_2=(ln(3))/(0.15) = 2.99 \approx 3\;h`