Question

If a city with a population of 52,000 doubles in size every 97 years, what will the population be 291 years from now?

Answer 1

To determine the population of a city that doubles in size every 97 years, we can use the formula for exponential growth. The population \( P \) at a given time can be expressed as:

\[
P(t) = P_0 \times 2^{(t/T)}
\]

where:
- \( P_0 \) is the initial population,
- \( t \) is the time elapsed,
- \( T \) is the time it takes for the population to double (97 years in this case).

Given:
- Initial population \( P_0 = 52,000 \),
- Time elapsed \( t = 291 \) years,
- Doubling time \( T = 97 \) years.

Now, we can calculate the number of doubling periods in 291 years:

\[
\text{Number of doublings} = \frac{t}{T} = \frac{291}{97} \approx 3
\]

Since \( 291 \div 97 \) equals approximately 3.00, indicating that there will be 3 complete doubling periods in 291 years.

Now, we apply this to our population formula:

\[
P(291) = 52,000 \times 2^{3}
\]

Calculating \( 2^{3} \):

\[
2^{3} = 8
\]

Next, we can find the population after 291 years:

\[
P(291) = 52,000 \times 8 = 416,000
\]

Thus, the population of the city in 291 years will be **416,000**.