Question

A colony of 55,700 bacteria doubles in size every 163 minutes. What will the population be 978 minutes from now?

Answer 1

To find out the population of the bacteria after 978 minutes, we can use the concept of exponential growth due to doubling.

1. **Determine the number of doubling periods in 978 minutes:**
The time taken for one doubling is 163 minutes. To find out how many times the bacteria will double in 978 minutes, we divide 978 by 163:

\[
\text{Number of doublings} = \frac{978}{163} \approx 6
\]

We can confirm this is approximately 6 as:

\[
163 \times 6 = 978 \quad \text{(exactly)}
\]

2. **Calculate the population after 6 doublings:**
The population doubles every time, so the formula for the population after \( n \) doublings is:

\[
P = P_0 \times 2^n
\]

where:
- \( P_0 = 55,700 \) (initial population),
- \( n = 6 \) (number of doublings).

Now substituting these values:

\[
P = 55,700 \times 2^6
\]

We first calculate \( 2^6 \):

\[
2^6 = 64
\]

Now, substitute this back into the equation:

\[
P = 55,700 \times 64
\]

3. **Perform the multiplication:**

\[
P = 55,700 \times 64 = 3,564,800
\]

Therefore, the population of the bacteria after 978 minutes will be **3,564,800**.