Question

(5 points each for a and b) A bacteria colony starts with 20 bacteria andgrows continuously at a rate of 28% per hour.a. How long will it take for the colony to:i. Double its size?ii. Reach 500,000 bacteria?b. How many bacteria will there be in:i. 3 hours?ii. 3.5 days?

Answer 1

Given:

The initial population of bacteria, I = 20

Growth rate, r = 28%

Step-by-step explanation:

a) To find: The time

i) Double its size

Using the formula,

`F=I(1+r)^t,\text{ Where I denotes initial and F denotes Final size.}`

On substitution we get,

`\begin{gathered} 40=20(1+0.28)^t \\ 1.28^t=(40)/(20) \\ t=\log _(1.28)2 \\ t=2.81 \end{gathered}`

Thus, the answer is 2.81 hours.

ii) To reach 500000 bacteria:

`\begin{gathered} 500000=20(1+0.28)^t \\ 1.28^t=(500000)/(20) \\ t=\log _(1.28)(25000) \\ t=41.02\text{ hours} \end{gathered}`

Thus, the answer is 41.02 hours.

b) To find the bacteria size:

i) In 3 hours,

`\begin{gathered} F=20(1+0.28)^3 \\ =41.94304 \\ \approx42 \end{gathered}`

Thus, the size of bacteria in 3 hours is 42.

i) In 3.5 days,

That is, 84 hours

`\begin{gathered} F=20(1+0.28)^(84) \\ =20261306488.67 \\ \approx20261306489 \end{gathered}`

Thus, the size of bacteria in 3.5 days is 20261306489.