Suppose a certain species of bacteria doubles in population every hour. If the initial population is 1,000 bacteria, find the number of bacteria at the 6th hour.

To solve this problem, we can use the formula for exponential growth, where the population doubles every hour:

\[
P(t) = P_0 \times 2^t
\]

Where:
- \( P(t) \) is the population at time \( t \) (in hours),
- \( P_0 \) is the initial population,
- \( t \) is the number of hours.

Given:
- \( P_0 = 1000 \) (initial population),
- \( t = 6 \) hours.

Now, substituting the values into the formula:

\[
P(6) = 1000 \times 2^6
\]

First, calculate \( 2^6 \):

\[
2^6 = 64
\]

Now, multiply:

\[
P(6) = 1000 \times 64 = 64,000
\]

So, the number of bacteria at the 6th hour is **64,000**.

Would you like more details or have any questions? Here are some related questions:

1. What would the population be at the 10th hour?
2. How long would it take for the population to exceed 1 million?
3. If the bacteria doubled every 30 minutes, how would that change the result?
4. What is the population at the 4th hour?
5. How would the population grow if it tripled every hour instead of doubling?

**Tip:** In exponential growth problems, understanding the growth factor (e.g., doubling, tripling) is key to setting up the correct formula.

Solve the problem of bacteria population growth where the population doubles every hour. Learn how to calculate the population at the 6th hour with an initial population of 1,000 bacteria. This exponential growth problem is suitable for students in Grades 6-8.

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