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:
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Initial Population: At \( t = 0 \): \[ n(0) = 50(1.2)^0 = 50 \] This means the initial population of the bacteria is 50.
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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 (\( t = 2 \)): \[ n(2) = 50(1.2)^2 = 50 \times 1.44 = 72 \]
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Exponential Growth: Since the growth is exponential due to the term \( (1.2)^t \), the population will increase rapidly as time goes on.
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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.
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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!