Answer :
To solve this problem, we utilize the exponential growth formula given by \( P = A e^{kt} \), where:
- \(P\) is the final population of bacteria,
- \(A\) is the initial population of bacteria,
- \(k\) is the growth rate constant,
- \(t\) is the time in hours.
Step-by-step solution:
1. Identify the given values:
- Initial bacteria (\(A\)) = 4,000
- Bacteria after 7 hours (\(P_{7}\)) = 5,200
- Time (\(t\)) = 7 hours
2. Using the formula \( P = A e^{kt} \), we can set up the equation for the growth rate:
[tex]\[ 5200 = 4000 e^{7k} \][/tex]
3. Solve for \(k\):
- Divide both sides by 4000:
[tex]\[ \frac{5200}{4000} = e^{7k} \][/tex]
[tex]\[ 1.3 = e^{7k} \][/tex]
- Take the natural logarithm of both sides:
[tex]\[ \ln(1.3) = 7k \][/tex]
- Solve for \(k\):
[tex]\[ k = \frac{\ln(1.3)}{7} \][/tex]
- Calculate \(k\):
[tex]\[ k \approx 0.0375 \quad \text{(rounded to four decimal places)} \][/tex]
4. Predict the number of bacteria after 9 hours. Use the same exponential growth formula with \(t = 9\):
[tex]\[ P_{9} = 4000 e^{0.0375 \times 9} \][/tex]
5. Calculate \(P_{9}\):
[tex]\[ P_{9} = 4000 e^{0.3375} \][/tex]
- Compute \(e^{0.3375}\):
[tex]\[ e^{0.3375} \approx 1.4013 \][/tex]
- Multiply by the initial population:
[tex]\[ P_{9} = 4000 \times 1.4013 \approx 5605 \][/tex]
After rounding the final result to the nearest whole number, we predict that the number of bacteria present after 9 hours is:
[tex]\[ \boxed{5605} \][/tex]
- \(P\) is the final population of bacteria,
- \(A\) is the initial population of bacteria,
- \(k\) is the growth rate constant,
- \(t\) is the time in hours.
Step-by-step solution:
1. Identify the given values:
- Initial bacteria (\(A\)) = 4,000
- Bacteria after 7 hours (\(P_{7}\)) = 5,200
- Time (\(t\)) = 7 hours
2. Using the formula \( P = A e^{kt} \), we can set up the equation for the growth rate:
[tex]\[ 5200 = 4000 e^{7k} \][/tex]
3. Solve for \(k\):
- Divide both sides by 4000:
[tex]\[ \frac{5200}{4000} = e^{7k} \][/tex]
[tex]\[ 1.3 = e^{7k} \][/tex]
- Take the natural logarithm of both sides:
[tex]\[ \ln(1.3) = 7k \][/tex]
- Solve for \(k\):
[tex]\[ k = \frac{\ln(1.3)}{7} \][/tex]
- Calculate \(k\):
[tex]\[ k \approx 0.0375 \quad \text{(rounded to four decimal places)} \][/tex]
4. Predict the number of bacteria after 9 hours. Use the same exponential growth formula with \(t = 9\):
[tex]\[ P_{9} = 4000 e^{0.0375 \times 9} \][/tex]
5. Calculate \(P_{9}\):
[tex]\[ P_{9} = 4000 e^{0.3375} \][/tex]
- Compute \(e^{0.3375}\):
[tex]\[ e^{0.3375} \approx 1.4013 \][/tex]
- Multiply by the initial population:
[tex]\[ P_{9} = 4000 \times 1.4013 \approx 5605 \][/tex]
After rounding the final result to the nearest whole number, we predict that the number of bacteria present after 9 hours is:
[tex]\[ \boxed{5605} \][/tex]
