Question

A scientist is observing a particular species of bacteria growing in a petri dish. The number of bacteria can be represented by the equatioN n (t) = 50(1.2)^t, there n is the number of bacteria after t hours

Answer 1

The equation you have is \( n(t) = 50(1.2)^t \), where:

- \( n(t) \) represents the number of bacteria at time \( t \) (in hours).
- The initial quantity of bacteria is 50 (when \( t = 0 \)).
- The growth factor is 1.2, indicating that the bacteria population grows by 20% every hour.

From this equation, you can analyze various aspects of the bacteria population:

1. **Initial Population**: At \( t = 0 \):
\[
n(0) = 50(1.2)^0 = 50
\]
This means the initial population of the bacteria is 50.

2. **Population After Specific Time**: To find the number of bacteria after a certain number of hours, simply substitute \( t \) with that number. For example:
- After 1 hour (\( t = 1 \)):
\[
n(1) = 50(1.2)^1 = 50 \times 1.2 = 60
\]
- After 2 hours (\( t = 2 \)):
\[
n(2) = 50(1.2)^2 = 50 \times 1.44 = 72
\]

3. **Exponential Growth**: Since the growth is exponential due to the term \( (1.2)^t \), the population will increase rapidly as time goes on.

4. **Long-Term Growth**: You can determine the population at larger values of \( t \) as well. For larger \( t \), the bacteria population will continue to grow without bound.

5. **Doubling Time**: To find how long it takes for the population to double, you can solve for \( t \) when \( n(t) = 100 \):
\[
100 = 50(1.2)^t
\]
Simplifying this gives:
\[
2 = (1.2)^t
\]
Taking the logarithm of both sides:
\[
t = \frac{\log(2)}{\log(1.2)} \approx 3.8 \text{ hours}
\]

This illustrates how you can analyze the growth of the bacterial population using the given equation. If you have specific questions or need calculations for other values of \( t \), feel free to ask!