Alright, let's analyze the given information step by step and derive the correct equation for the number of bacteria after [tex]\(x\)[/tex] days.
Firstly, we need to understand how the bacteria population grows:
1. Initial Population: The experiment starts with an initial population of 500 bacteria.
2. Doubling Time: The population doubles every 4 hours.
Now, let’s break down the time period:
- There are 24 hours in a day.
- Since the population doubles every 4 hours, we need to determine how many times the population doubles in one day.
Calculating the number of doublings in one day:
- In one day (24 hours), how many 4-hour periods are there?
[tex]\[
\text{Number of 4-hour periods in a day} = \frac{24 \text{ hours}}{4 \text{ hours/period}} = 6
\][/tex]
Therefore, the population doubles 6 times in one day.
Let’s denote the number of days by [tex]\(x\)[/tex].
3. Forming the Equation:
- After [tex]\(x\)[/tex] days, the total number of 4-hour periods is [tex]\(6x\)[/tex], since there are 6 periods in each day.
- The population doubles every 4 hours, so after [tex]\(6x\)[/tex] periods, the population will be:
[tex]\[
n = 500 \times 2^{6x}
\][/tex]
Hence, the equation that gives the expected number of bacteria [tex]\(n\)[/tex] after [tex]\(x\)[/tex] days is:
[tex]\[
n = 500 \times 2^{6x}
\][/tex]
Now we can identify which of the given equations matches this form:
A. [tex]\( n = 500 \times 2^x \)[/tex] — Incorrect, as it does not account for the [tex]\(6\)[/tex] doublings per day.
B. [tex]\( n = 500 \times 2^{6 x} \)[/tex] — Correct, aligns with our derived formula.
C. [tex]\( n = 500 \times 6^x \)[/tex] — Incorrect, as it implies multiplying by 6 per day, which is not the case.
D. [tex]\( n = 500 \times 6^{2 x} \)[/tex] — Incorrect, as it also does not match the derived exponent pattern.
Therefore, the correct equation is:
B. [tex]\( n = 500 \times 2^{6x} \)[/tex]
