Math Problem Statement

The count in a bacteria culture was 200 after 20 minutes and 1900 after 35 minutes. Assuming the count grows exponentially, What was the initial size of the culture? Find the doubling period. Find the population after 120 minutes. When will the population reach 11000. You may enter the exact value or round to 2 decimal places.

Solution

We are given that the growth of the bacteria culture is exponential. The general form for exponential growth is:

P(t)=P0ektP(t) = P_0 e^{kt} where:

  • P(t)P(t) is the population at time tt,
  • P0P_0 is the initial population (what we're solving for),
  • kk is the growth rate constant,
  • tt is the time in minutes,
  • ee is Euler's number (approximately 2.718).

Step 1: Set up the equations from the given information

We know the following:

  • At t=20t = 20 minutes, the population is 200: P(20)=200=P0e20kP(20) = 200 = P_0 e^{20k}
  • At t=35t = 35 minutes, the population is 1900: P(35)=1900=P0e35kP(35) = 1900 = P_0 e^{35k}

Step 2: Solve for kk

To find kk, divide the second equation by the first to eliminate P0P_0:

1900200=P0e35kP0e20k\frac{1900}{200} = \frac{P_0 e^{35k}}{P_0 e^{20k}}

9.5=e15k9.5 = e^{15k}

Now take the natural logarithm of both sides:

ln(9.5)=15k\ln(9.5) = 15k

k=ln(9.5)152.2513150.15009k = \frac{\ln(9.5)}{15} \approx \frac{2.2513}{15} \approx 0.15009

Step 3: Solve for P0P_0

Substitute k0.15009k \approx 0.15009 into one of the original equations to find P0P_0. Let's use the equation for t=20t = 20:

200=P0e20(0.15009)200 = P_0 e^{20(0.15009)}

200=P0e3.0018200 = P_0 e^{3.0018}

P0=200e3.001820020.19.95P_0 = \frac{200}{e^{3.0018}} \approx \frac{200}{20.1} \approx 9.95

So, the initial population P0P_0 is approximately 9.95.

Step 4: Find the doubling period

The doubling period TdT_d is the time it takes for the population to double. It can be calculated using:

Td=ln(2)kT_d = \frac{\ln(2)}{k}

Substitute k0.15009k \approx 0.15009:

Td=ln(2)0.150090.69310.150094.62 minutesT_d = \frac{\ln(2)}{0.15009} \approx \frac{0.6931}{0.15009} \approx 4.62 \text{ minutes}

Step 5: Find the population after 120 minutes

Use the formula P(t)=P0ektP(t) = P_0 e^{kt} to find the population at t=120t = 120:

P(120)=9.95e0.15009×120P(120) = 9.95 e^{0.15009 \times 120}

P(120)=9.95e18.01089.95×660807.36570954.67P(120) = 9.95 e^{18.0108} \approx 9.95 \times 660807.3 \approx 6570954.67

So, the population after 120 minutes is approximately 6,570,954.67.

Step 6: Find when the population will reach 11,000

To find when the population reaches 11,000, we solve:

11000=9.95e0.15009t11000 = 9.95 e^{0.15009t}

Divide both sides by 9.95:

1105.53=e0.15009t1105.53 = e^{0.15009t}

Take the natural logarithm of both sides:

ln(1105.53)=0.15009t\ln(1105.53) = 0.15009t

t=ln(1105.53)0.150097.0080.1500946.68 minutest = \frac{\ln(1105.53)}{0.15009} \approx \frac{7.008}{0.15009} \approx 46.68 \text{ minutes}

So, the population will reach 11,000 after approximately 46.68 minutes.


Summary of Results:

  1. Initial population P0P_0: 9.95
  2. Doubling period TdT_d: 4.62 minutes
  3. Population after 120 minutes: 6,570,954.67
  4. Time to reach a population of 11,000: 46.68 minutes

Would you like to explore any of these calculations further? Here are 5 related questions:

  1. How does the growth constant kk affect the doubling period?
  2. Can we use this method for other exponential growth scenarios, like radioactive decay?
  3. What would the population be after 240 minutes?
  4. What if the bacteria grew at a slower rate, say half the current growth constant?
  5. How would a change in the initial population affect the results?

Tip: When solving exponential growth problems, always check if you can simplify by taking logarithms at the right step!

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Logarithms

Formulas

Exponential growth formula: P(t) = P_0 * e^(kt)
Doubling period formula: T_d = ln(2) / k
Logarithmic formula for solving growth rate: ln(P2 / P1) = k * (t2 - t1)

Theorems

Exponential Growth Theorem

Suitable Grade Level

Grades 10-12