If a town with a population of 10,000 doubles in size every 17 years, what will the population be 68 years from now?
Understand the Problem
The question is asking us to calculate the population of a town after 68 years, given that the population doubles every 17 years starting from 10,000. We will determine how many doubling periods fit into 68 years and then compute the resulting population.
Answer
The population will be $160,000$ after 68 years.
Answer for screen readers
The population of the town after 68 years will be $160,000$.
Steps to Solve
- Determine the number of doubling periods To find out how many times the population doubles in 68 years, we can divide 68 by the doubling period of 17 years.
$$ \text{Number of doublings} = \frac{68}{17} = 4 $$
- Calculate the new population Starting with an initial population of 10,000, we calculate the population after 4 doubling periods. Each doubling means the population multiplies by 2.
$$ \text{Population after 4 doublings} = 10,000 \times 2^4 $$
- Solve the population calculation Now, calculate the value of $2^4$:
$$ 2^4 = 16 $$
So, we substitute this back into our population equation:
$$ \text{Population} = 10,000 \times 16 $$
- Final calculation Now calculate the final population:
$$ \text{Population} = 160,000 $$
The population of the town after 68 years will be $160,000$.
More Information
The growth of the town's population showcases exponential growth, which is common in biological systems and can occur in various populations under ideal conditions. The observation of doubling in fixed intervals is a foundational concept in population studies.
Tips
- Miscounting the number of doublings: Ensure that the timeframe is correctly divided by the doubling period.
- Forgetting to multiply properly: It's essential to remember to raise 2 to the power corresponding to the number of doublings before multiplying by the initial population.
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