The count in a bacteria culture was 500 after 20 minutes and 1600 after 40 minutes. Assuming the count grows exponentially,

Question

The count in a bacteria culture was 500 after 20 minutes and 1600 after 40 minutes. Assuming the count grows exponentially,

What was the initial size of the culture?

Find the doubling period.

Find the population after 105 minutes.

When will the population reach 14000.

pls help and explain



Answer 1

Pop(t)= Pop(0)2^t/T where Pop(t) is the population as a function of time and T is the doubling time constant.

Given that Pop(40) = Pop(0)(2^20/T)²=1600 and Pop(0)e^20/T=500, 2^20/T = 500/Pop(0). Solving 1600=Pop(0)(500/Pop(0))² yields Pop(0) = 156.25. Now 2^20/T=500/156.25 yields T= 11.9 minutes.

Pop(105)=156.2*(2)^105/11.9 = 70766

Pop(t)=14000=156.2*(2)^t/11.9 yields t = 19.8 minutes

Answer 2

P=Ae^kt

500 = Ae^20k

1600 = Ae^40k

Divide the second equation by the first

3.2 = e^20k

ln(3.2) = 20k --> k = ln(3.2)/20

Therefore P = A(3.2)^t/20 since e^ln(3.2) _{= 3.2}

¹⁾

500 = A(3.2)^20/20 = 3.2A --> A = 156.25 Initial size

P(t) = 156.25(3.2)^t/20

2) To solve for t take log of both sides and isolate t

t = 20ln(2)/ln(3.2) = 11.91 minutes

3) P(105) = 156.25(3.2)^105/20 = 70122

4) 14000 = 156.25(3.2)^t/20

Take log of both sides

ln(14000/156.25) = (t/20) ln(3.2)

t = 20ln(89.6)/ln(3.2)

t= 77.296 minutes = 1 hour 17 minutes 17.76 seconds