The count in a bacteria culture was 400 after 10 minutes and 1500 after 40 minutes. Assuming the count grows exponentially,

Question

What was the initial size of the culture? bacteria

Find the doubling period. minutes

Find the population after 60 minutes. bacteria

When will the population reach 13000. minutes

Answer 1

We can use this form of exponential function:

A = Pe^rt

Where:

A=Final amount of bacteria

P=The initial amount or size of bacteria

r= rate

t = Time

400 = Pe^10r

1500 = Pe^40r

From the first equation:

A = 400/e^10r

Plugin to the second equations:

1500 = (400/e^10r)e^40r

1500 = 400e^30r

1500/400=e^30r

15/4 =e^30r

ln (15/4) = 30r

r = ln (15/4)/30

(1)Solve for P:

400 = Pe^{10(ln(15/4)/30)}

400 = Pe^{(1/3)ln(15/4)}

400 = P(15/4)^(1/3)

400/(15/4)^(1/3)=P

P ≈ 257.47≈ 257 bacteria

(2)Find the doubling period:

2 = (15/4)^(t/30)

ln(2) = (t/30)•ln(15/4)

ln(2)/ln(15/4) = t/30

t = 30*ln(2)/ln(15/4) ≈ 15.73 mins.

(3)Our function is: A(t) = 257.47(15/4)^t/30

A= 257.47(15/4)^60/30)

A≈ 3621 Bacteria

(4)13000=257.47(15/4)^t/30

13000/257.47 = (15/4)^t/30

ln(13000/257.47) = (t/30)ln(15/4)

ln(13000/257.47)/ln(15/4) =t/30

t = 30•ln(13000/257.47)/ln(15/4)

t = 89.01 mins.

*Note: I just round off on the final answer to make it accurate.

Answer 2

400 = a₀e¹

1500 = a₀e^40k

3.75 = e^30k

ln 3.75 = 30k

1.322 ≈ 30k

0.044 ≈ k

2 = e^0.044t

ln 2 = 0.044t

0.693 ≈ 0.044t

15.753 ≈ t

Can you compute the next two and answer?