MATH SOLVE

2 months ago

Q:
# 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?

Accepted Solution

A:

Answer: a) [tex]N(t) = 10^3\exp(0.046\frac{1}{min}t)[/tex]b) 1,000,000 bacteria at t = 150 minStep-by-step explanation:Hi!!A colony that grows exponentially has a number of bacteria:[tex]N(t) = N_0 \exp(\lambda t)[/tex]In this case at time t = 0:[tex]N(0)=N_0=10^3[/tex]We need to find the value of λ. We use the data:[tex]N(t=50\;min)10^4 = 10^3\exp(\lambda \;50\;min)[/tex][tex]ln(10)=2.3=\lambda\;50\;min\\\lambda= \frac{0.046}{min}\\N(t) = 10^3\exp(\frac{0.046}{min}t)\\[/tex]To find when there will be 1,000,000 bacteria:[tex]10^6=10^3\exp(\frac{0.046}{min}t)[/tex][tex]\ln(10^3)=3\ln(10) = \frac{0.046}{min}t[/tex][tex]t = 150\;min [/tex]