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Difficulty: EasyExponential Functions and Equations

A population of bacteria doubles every 3 hours. If the initial population of the bacteria is 500, what is the population of the bacteria after 9 hours?

Answer: 4000 bacteria

Answer

The population of the bacteria after 9 hours is 4,000.
The final population is calculated using the formula P(t)=P0×2t/dP(t) = P_0 \times 2^{t/d}, where P0P_0 is the initial population of 500, dd is the doubling period of 3 hours, and tt is the total time of 9 hours. Evaluating this gives P(9)=500×29/3=500×23=500×8=4000P(9) = 500 \times 2^{9/3} = 500 \times 2^3 = 500 \times 8 = 4000.

Step-by-Step Solution

1
Identify the initial population (P0P_0), doubling time (dd), and total time (tt).
P0=500P_0 = 500, d=3d = 3, and t=9t = 9.
These parameters are required to set up the exponential growth model.
2
Calculate the number of doubling periods.
The number of doubling periods is 93=3\frac{9}{3} = 3.
The population doubles once for every 3-hour interval.
3
Calculate the final population using the exponential growth formula.
500×23=500×8=4000500 \times 2^3 = 500 \times 8 = 4000.
Applying the 3 doubling cycles to the initial population of 500 yields the final population.

Key Concept

Exponential Growth Model
Estimated Time:1m 0s
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Topics

Exponential Functions and Equations

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