Question

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 500 bacteria are present. After 10 hours 4000 bacteria are present. What was the initial number of bacteria? (Round your answer to the nearest integer.)

Answer 1

The initial number of bacteria is 371.

Step-by-step explanation:

Let P represents population of the bacteria and t represents time,

According to the question,

`(dP)/(dt)\propto P`

`\implies (dP)/(dt)=kP`

Where, k is constant of proportionality,

`(dP)/(dt)=kdt`

Integrating both sides,

`ln P=kt+C`

`P=e^(kt+C)`

`P=e^C e^(kt)`

`P=P_0 e^(kt)` ( Let
`P_0=e^C` )

If t = 0,
`P=P_0`

That is,
`P_0` is the intial population.

Since, when t = 3, P = 500 and when t = 10, P=4000

`500=P_0 e^(3t)`

`4000 = P_0 e^(10t)`

`\because (4000)/(500)=( P_0 e^(10t))/( P_0 e^(3t))`

`8=e^((10-3)t)=e^(7t)\implies 7t = ln(8) \implies t=(ln(8))/(7)=0.297063077383\approx 0.2971`

`\implies 500=P_0 e^(0.2971)\implies P_0=(500)/(e^(0.2971))= 371.484855838\approx 371`

Hence, initial population is approximately 371.

Answer 2

The initial number of bacteria was 7 bacterial cells.

What was the initial number of bacteria?

P (t) = P 0 and rt

Where:

P(t) = the amount of some quantity at time t

P 0 = initial amount at time t = 0

r = growth rate

t = time (number of periods)

Initial amount (P 0 ) 7.21940555138e-32

Growth rate (r) 8 = 800.0%

Time (t) 10

Final amount (P (t)) 4000

With this information, we can conclude that the initial rounded number of bacterial cells is 7.