Question

Consider a colony of E.Coli bacteria that is growing exponentially. A microbiologist finds that, initially, 1,000 bacteria are present and 50 minutes later there are 10,000 bacteria. a) Find expression for the number of bacteria Q(t) after t minutes. b) When will there be 1,000,000 bacteria?

Answer 1

Answer:

a)$Q(t)=Q_{0}e^{(ln10/50)t}$

b) 150min

Step-by-step explanation:

a) Using the formula for colony growth $Q(t)=Q_{0}e^{kt}$ we need to find the specific value of k, to do this you'll use the time, initial and final number of colonies provided in your problem.

$Q(t)=Q_{0}e^{kt}\\\\10000=1000e^{k(50)} \\\\\frac{10000}{1000}=e^{50k} \\\\10=e^{50k}\\ ln(10=e^{50k})\\\\ln10=50k\\\\k=ln10/50\\\\Q(t)=Q_{0}e^{(ln10/50)t}$

b) Once we have our expression we only have to use our final and initial number of bacteria to find t.

$Q(t)=Q_{0}e^{(ln10/50)t}\\\\\\1000000=1000e^{0.046t}\\ 1000=e^{0.046t}\\ln( 1000=e^{0.046t})\\ln1000=0.046t\\t=ln1000/0.046\\t=150 minutes$

I hope you find this information useful! good luck!