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Difficulty: EasyLinear and Exponential Growth

A scientist is monitoring the population of a bacterial culture in a petri dish. The population of the culture grows exponentially, doubling every 33 hours. If the population of the culture after 99 hours is 1,6001,600, what was the initial population of the culture?

Answer: 200

Answer

The initial population of the culture was 200200.
Since the bacterial culture doubles every 33 hours, the population undergoes 9÷3=39 \div 3 = 3 doubling periods over a span of 99 hours. An initial population P0P_0 that doubles 33 times will grow to P0×23=8P0P_0 \times 2^3 = 8P_0. Setting this expression equal to the final population of 1,6001,600 gives 8P0=1,6008P_0 = 1,600. Dividing both sides of this equation by 88 yields P0=200P_0 = 200. Therefore, the initial population of the culture was 200200.

Step-by-Step Solution

1
Determine the number of doubling periods that occurred in 99 hours.
33 doubling periods
Since the population doubles every 33 hours, dividing the total time of 99 hours by the doubling time of 33 hours gives 9÷3=39 \div 3 = 3 periods.
2
Set up an equation representing the exponential growth.
P0×23=1,600P_0 \times 2^3 = 1,600
An initial population P0P_0 doubling 33 times grows by a factor of 232^3, which is equal to 88 times the initial amount.
3
Solve the equation 8P0=1,6008P_0 = 1,600 for the initial population P0P_0.
P0=200P_0 = 200
Dividing both sides of the equation by 88 isolates P0P_0 and gives the initial value.

Key Concept

Exponential growth models and solving for the initial value.
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Topics

Linear and Exponential Growth

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