Question

question 15
the count in a bacteria culture was 100 after 10 minutes and 1200 after 30 minutes. assuming the count grows exponentially,
what was the initial size of the culture? □ bacteria
find the doubling period. □ minutes
find the population after 75 minutes. □ bacteria
when will the population reach 15000. □ minutes

Explanation:

Step1: Define Exponential Growth Model

The exponential growth formula is \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, and \( k \) is the growth rate constant. We know \( P(10) = 100 \) and \( P(30) = 1200 \).

Step2: Set Up Equations

For \( t = 10 \): \( 100 = P_0 e^{10k} \)
For \( t = 30 \): \( 1200 = P_0 e^{30k} \)

Step3: Divide the Second Equation by the First

\( \frac{1200}{100} = \frac{P_0 e^{30k}}{P_0 e^{10k}} \)
Simplify: \( 12 = e^{20k} \)

Step4: Solve for \( k \)

Take natural log: \( \ln(12) = 20k \)
\( k = \frac{\ln(12)}{20} \approx \frac{2.4849}{20} \approx 0.1242 \)

Step5: Find \( P_0 \)

Substitute \( k \) into \( 100 = P_0 e^{10k} \):
\( 100 = P_0 e^{10 \times 0.1242} \)
\( 100 = P_0 e^{1.242} \)
\( e^{1.242} \approx 3.46 \)
\( P_0 = \frac{100}{3.46} \approx 28.9 \approx 29 \) (or more accurately, use exact terms: from \( 12 = e^{20k} \), \( e^{10k} = \sqrt{12} = 2\sqrt{3} \), so \( P_0 = \frac{100}{2\sqrt{3}} = \frac{50}{\sqrt{3}} \approx 28.87 \approx 29 \))

Step6: Find Doubling Period (\( T \))

Doubling when \( P(t) = 2P_0 \): \( 2P_0 = P_0 e^{kT} \)
Simplify: \( 2 = e^{kT} \)
Take log: \( \ln(2) = kT \)
\( T = \frac{\ln(2)}{k} = \frac{\ln(2) \times 20}{\ln(12)} \approx \frac{0.6931 \times 20}{2.4849} \approx \frac{13.862}{2.4849} \approx 5.58 \) minutes

Step7: Population at \( t = 75 \)

\( P(75) = P_0 e^{75k} \)
Substitute \( P_0 = \frac{50}{\sqrt{3}} \) and \( k = \frac{\ln(12)}{20} \):
\( P(75) = \frac{50}{\sqrt{3}} e^{75 \times \frac{\ln(12)}{20}} = \frac{50}{\sqrt{3}} e^{\frac{15}{4} \ln(12)} = \frac{50}{\sqrt{3}} (12)^{\frac{15}{4}} \)
Calculate \( 12^{\frac{15}{4}} = 12^{3.75} = (12^3)(12^{0.75}) = 1728 \times \sqrt[4]{12^3} = 1728 \times \sqrt[4]{1728} \approx 1728 \times 6.96 \approx 12047 \), then \( \frac{50}{\sqrt{3}} \times 12047 \approx 28.87 \times 12047 \approx 347,800 \) (or use \( P_0 \approx 28.87 \), \( k \approx 0.1242 \)):
\( P(75) = 28.87 e^{0.1242 \times 75} = 28.87 e^{9.315} \approx 28.87 \times 10950 \approx 316,000 \)? Wait, miscalculation earlier. Let's recalculate \( k \): \( \ln(12) \approx 2.48490665 \), so \( k = 2.48490665 / 20 = 0.12424533 \). Then \( 75k = 75 \times 0.12424533 = 9.3184 \). \( e^{9.3184} \approx e^9 \times e^{0.3184} \approx 8103.08 \times 1.375 \approx 11142 \). Then \( P(75) = 28.87 \times 11142 \approx 321,000 \)? Wait, better to use the ratio. From \( t=10 \) to \( t=30 \) (20 minutes), population multiplies by 12. From \( t=30 \) to \( t=75 \) is 45 minutes, which is \( 45/20 = 2.25 \) intervals of 20 minutes. So multiplier is \( 12^{2.25} = 12^{9/4} = (12^{1/4})^9 \). \( 12^{1/4} \approx 1.861 \), so \( 1.861^9 \approx 1.861^2=3.463, ^4=11.99, ^8=143.7, ^9=143.7×1.861≈267.4 \). Then \( P(75) = 1200 \times 267.4 ≈ 320,880 \). Wait, maybe my initial \( P_0 \) was approximated. Let's use exact \( P_0 = 100 / e^{10k} \), and \( e^{30k} = 12 e^{10k} \), so \( P(75) = P_0 e^{75k} = 100 e^{65k} = 100 (e^{20k})^{3.25} = 100 (12)^{3.25} = 100 \times 12^3 \times 12^{0.25} = 100 \times 1728 \times \sqrt[4]{12} ≈ 100 \times 1728 \times 1.861 ≈ 100 \times 3216 ≈ 321600 \).

Step8: Time to Reach 15000

\( 15000 = P_0 e^{kt} \)
\( 15000 = \frac{50}{\sqrt{3}} e^{kt} \)
\( e^{kt} = \frac{15000 \sqrt{3}}{50} = 300 \sqrt{3} ≈ 519.6 \)
Take log: \( kt = \ln(519.6) ≈ 6.25 \)
\( t = \frac{6.25}{k} = \frac{6.25 \times 20}{\ln(12)} ≈ \frac{125}{2.4849} ≈ 50.3 \) minutes? Wait, no: \( kt = \ln(15000 / P_0) \). \( P_0 ≈ 28.87 \), so \( 15000 / 28.87 ≈ 519.6 \), \( \ln(519.6) ≈ 6.25 \), \( k ≈ 0.1242 \), so \( t = 6.25 / 0.1242 ≈ 50.3 \)? Wait, that can't be, because at t=30, it's 1200, at t=50, it's 1200 * e^{20k} = 1200*12=14400, at t=50.3, it's ~15000. Yes, because from t=30 (1200) to t=50 (20 minutes), it's 1200*12=14400, then a bit more. So \( t = 30 + \frac{\ln(15000/1200)}{k} = 30 + \frac{\ln(12.5)}{0.1242} ≈ 30 + \frac{2.5257}{0.1242} ≈ 30 + 20.34 ≈ 50.34 \) minutes.

Correcting Initial \( P_0 \) (More Accurate)

From \( 100 = P_0 e^{10k} \) and \( 1200 = P_0 e^{30k} \), divide the second by the first: \( 12 = e^{20k} \implies e^{10k} = \sqrt{12} = 2\sqrt{3} \approx 3.4641 \). Thus, \( P_0 = 100 / (2\sqrt{3}) = 50 / \sqrt{3} \approx 28.8675 \approx 29 \).

Doubling Period (More Accurate)

\( T = \ln(2)/k = \ln(2) / (\ln(12)/20) = 20 \ln(2) / \ln(12) \approx 20 * 0.6931 / 2.4849 \approx 13.862 / 2.4849 \approx 5.58 \) minutes.

Population at 75 Minutes

\( P(75) = P_0 e^{75k} = (50/\sqrt{3}) e^{75k} \). Since \( e^{20k} = 12 \), \( e^{75k} = e^{20k*3 + 15k} = 12^3 * e^{15k} \). \( e^{15k} = (e^{10k})^{1.5} = (2\sqrt{3})^{1.5} = (2\sqrt{3}) * (2\sqrt{3})^{0.5} = 2\sqrt{3} * \sqrt{2\sqrt{3}} = 2\sqrt{3} * (2^{0.5} * 3^{0.25}) = 2^{1.5} * 3^{0.75} = 2\sqrt{2} * 3^{3/4} \approx 2.828 * 2.2795 \approx 6.45 \). Then \( 12^3 = 1728 \), so \( e^{75k} \approx 1728 * 6.45 \approx 11145.6 \). Then \( P(75) = (50/\sqrt{3}) * 11145.6 \approx 28.8675 * 11145.6 \approx 321,000 \) (approximate).

Time to Reach 15000

\( 15000 = P_0 e^{kt} \implies t = \frac{\ln(15000 / P_0)}{k} = \frac{\ln(15000 / (50/\sqrt{3}))}{\ln(12)/20} = \frac{\ln(300\sqrt{3}) * 20}{\ln(12)} \). \( 300\sqrt{3} \approx 519.615 \), \( \ln(519.615) \approx 6.25 \), so \( t \approx \frac{6.25 * 20}{2.4849} \approx \frac{125}{2.4849} \approx 50.3 \) minutes.

Answers (Approximate):

- Initial size: \( \boldsymbol{\approx 29} \) bacteria
- Doubling period: \( \boldsymbol{\approx 5.58} \) minutes (or \( \approx 5.6 \))
- Population at 75 minutes: \( \boldsymbol{\approx 321000} \) bacteria (or more accurately, using exact steps: \( P(75) = 1200 \times 12^{(45/20)} = 1200 \times 12^{2.25} \approx 1200 \times 267.4 \approx 320,880 \))
- Time to reach 15000: \( \boldsymbol{\approx 50.3} \) minutes

(Note: Depending on rounding during steps, slight variations may occur. For example, using \( P_0 = 28.87 \), \( k = 0.1242 \), the calculations will align.)

Answer:

s (Approximate):

• Initial size: \( \boldsymbol{\approx 29} \) bacteria
• Doubling period: \( \boldsymbol{\approx 5.58} \) minutes (or \( \approx 5.6 \))
• Population at 75 minutes: \( \boldsymbol{\approx 321000} \) bacteria (or more accurately, using exact steps: \( P(75) = 1200 \times 12^{(45/20)} = 1200 \times 12^{2.25} \approx 1200 \times 267.4 \approx 320,880 \))
• Time to reach 15000: \( \boldsymbol{\approx 50.3} \) minutes

(Note: Depending on rounding during steps, slight variations may occur. For example, using \( P_0 = 28.87 \), \( k = 0.1242 \), the calculations will align.)