Question

The count in a bacteria culture was 800 after 15 minutes and 1000 after 30 minutes. Assuming the count grows exponentiallyA)What was the initial size of the culture? B)Find the doubling period. C)Find the population after 60 minutes. D)When will the population reach 13000.

Answer 1

A) The initial size o the culture is 640

B) The doubling period is 47 minutes

C) The population after 60 minutes is 1563

D) The population will reach 13000 after 3 hours 22 minutes

Step-by-step explanation:

The form of an exponential grow model is:

`S=Pb^t`

Where:

S is the population after t hours

P is the initial population

b is the base of the exponent

t is the time, in hours

We know that after 15 minutes, the population was 800. 15 minutes is a quarter of an hour. Thus, t = 1/4, S = 800:

`800=Pb^{(1)/(4)}`

Also, we know that after 30 minutes, the population was 1000. Thus, t = 1/2, S = 1000

`1000=Pb^{(1)/(2)}`

Then, we have a system of equations:

`\begin{cases}800=Pb^{(1)/(4)}{} \\ 1000=Pb^{(1)/(2)}{}\end{cases}`

We can solve the first equation for P:

`\begin{gathered} 800=Pb^{(1)/(4)} \\ P=\frac{800}{b^{(1)/(4)}} \end{gathered}`

And substitute in the other equation:

`1000=\frac{800}{b^{(1)/(4)}}b^{(1)/(2)}`

And solve:

`(1000)/(800)=\frac{b^{(1)/(2)}}{b^{(1)/(4)}}`
`\begin{gathered} (5)/(4)=b^{(1)/(2)-(1)/(4)} \\ . \\ (5)/(4)=b^{(1)/(4)} \end{gathered}`
`\begin{gathered} b=((5)/(4))^4 \\ . \\ b=(625)/(256) \end{gathered}`

Now, we can find the initial population P:

`P=(800)/(((625)/(256))^4)=(800)/((5)/(4))=(800\cdot4)/(5)=640`

The initial population is 640

To find the doubling period, we want that the population equal to twice the initial population:

`S=2P`

Then, since we know the equation, we can write:

`2P=P((625)/(256))^t`

Then:

`\begin{gathered} (2P)/(P)=((625)/(256))^t \\ . \\ 2=((625)/(256))^t \\ \ln(2)=t\ln((625)/(256)) \\ . \\ (\ln(2))/(\ln((625)/(256)))=t \\ . \\ t\approx0.7765 \end{gathered}`

If an hour is 60 minutes:

`60\cdot0.7765=46.59\approx47\text{ }minutes`

To find the population after 60 minutes, we use t = 1 hour and we want to find S:

`\begin{gathered} S=640((625)/(256))^1 \\ . \\ S=640\cdot(625)/(256)=1562.5 \end{gathered}`

To find when the population is 13000, then we use S = 13000 and solve for t:

`\begin{gathered} 13000=640((625)/(256))^t \\ . \\ (13000)/(640)=((625)/(256))^t \\ . \\ (325)/(16)=((625)/(256))^t \\ . \\ \ln((325)/(16))=t\ln((625)/(256))^ \\ . \\ t=(\ln((325)/(16)))/(\ln((625)/(256)))\approx3.373 \\ \\ \end{gathered}`

We have 3 full hours and 0.373. Since one hour is 60 minutes:

`60\cdot0.373\approx22`

The population reach 13000 after 3 hours 22 minutes