Bacteria usually reproduce by a process known as binary fission. In this type of reproduction, one bacterium divides, forming two bacteria. If the initial population is 100 bacteria, and the bacteria double every 8 hours, how many bacteria will there be in 3 days?

Set up and solve:

[tex]\[ 100 \times 2^{\left(\frac{72}{8}\right)} \][/tex]

(Note: 3 days is 72 hours.)

To solve this problem, let's break it down step by step:

1. Understand the Problem:
- We have an initial population of 100 bacteria.
- These bacteria double every 8 hours.
- We need to find out how many bacteria there will be after 3 days.

2. Convert Days into Hours:
- Since bacteria double every 8 hours, we need to work in hours.
- 3 days is equivalent to [tex]$3 \times 24 = 72$[/tex] hours.

3. Determine the Number of Doubling Periods:
- Bacteria double every 8 hours. Therefore, the number of 8-hour periods in 72 hours is calculated by dividing 72 by 8:
[tex]\[
\text{Number of Doublings} = \frac{72}{8} = 9
\][/tex]
- This means the bacteria will double 9 times in 3 days.

4. Calculate the Final Population:
- Every doubling multiplies the population by 2.
- We start with 100 bacteria, so after 9 doublings, the formula to find the final population is:
[tex]\[
\text{Final Population} = \text{Initial Population} \times 2^{\text{Number of Doublings}}
\][/tex]
- Plugging in the numbers:
[tex]\[
\text{Final Population} = 100 \times 2^9
\][/tex]

5. Perform the Calculation:
- Calculate [tex]$2^9 = 512$[/tex].
- Therefore, the final population is:
[tex]\[
100 \times 512 = 51200
\][/tex]

6. Conclusion:
- After 3 days, there will be 51,200 bacteria.

This step-by-step approach helps in understanding how the growth process continues over the given period, resulting in the number of bacteria reaching 51,200 at the end of 3 days.