Math Problem Statement

The population of a bacteria culture doubles every 3 hours. If the initial population is 100 bacteria, what will be the population after 9 hours?

Solution

The problem is asking for the population of a bacteria culture after 9 hours, given that the initial population is 100 bacteria and it doubles every 3 hours.

We can solve this using the formula for exponential growth:

P(t)=P0×2tTP(t) = P_0 \times 2^{\frac{t}{T}}

Where:

  • P(t)P(t) is the population after time tt,
  • P0P_0 is the initial population (100 bacteria),
  • tt is the time elapsed (9 hours),
  • TT is the doubling time (3 hours).

Now we can substitute the values:

P(9)=100×293=100×23=100×8=800P(9) = 100 \times 2^{\frac{9}{3}} = 100 \times 2^3 = 100 \times 8 = 800

Thus, the population after 9 hours will be 800 bacteria.

Would you like more details or have any questions? Here are some related questions for further understanding:

  1. What happens to the population if the doubling time changes to 2 hours?
  2. How would the formula change if the population increased by 50% every 3 hours instead of doubling?
  3. Can you use logarithms to find out how long it takes for the population to reach 1600 bacteria?
  4. How does exponential growth compare to linear growth in the context of population dynamics?
  5. What are some real-world examples where bacteria growth models like this are used?

Tip: Exponential growth often leads to very large populations quickly. It's important to know when exponential growth applies and when limiting factors slow the growth down (logistic growth).

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Doubling Time
Logarithms

Formulas

P(t) = P_0 * 2^(t/T)

Theorems

Exponential Growth Theorem

Suitable Grade Level

Grades 8-10