Answer:
a) [tex]N_0=250\; k=0.15 [/tex]
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 [tex]t=0[/tex]. And [tex]N(0)=N_0[/tex]. The process is:
[tex]N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250[/tex]
b) After 2 days means [tex]t=48[/tex]. So, we just replace and operate:
[tex]N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}[/tex]
c) [tex]N(t_1)=4000; \;t_1=?[/tex]
[tex]N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h [/tex]
d) [tex]t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)[/tex]
[tex]N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h [/tex]
