Answer :
To predict the number of bacteria present after 19 hours, let's follow a step-by-step solution using the given exponential growth formula:
[tex]\[ P = A e^{k t} \][/tex]
1. Initial Information:
- Initial number of bacteria (\(P_0\)) = 6,000
- Number of bacteria after 3 hours (\(P_3\)) = 7,200
- Time elapsed to reach 7,200 bacteria (\(t_3\)) = 3 hours
- Time after which we need to predict the number of bacteria (\(t_{19}\)) = 19 hours
2. Determine the growth rate constant \(k\):
Using the formula \(P = A e^{k t}\):
[tex]\[ P_3 = P_0 e^{k t_3} \][/tex]
Substitute \(P_3\), \(P_0\), and \(t_3\):
[tex]\[ 7200 = 6000 e^{3k} \][/tex]
Solve for \(e^{3k}\):
[tex]\[ e^{3k} = \frac{7200}{6000} \][/tex]
[tex]\[ e^{3k} = 1.2 \][/tex]
Take the natural logarithm on both sides to solve for \(3k\):
[tex]\[ 3k = \ln(1.2) \][/tex]
Divide by 3 to isolate \(k\):
[tex]\[ k = \frac{\ln(1.2)}{3} \][/tex]
Calculate \(k\) and round to four decimal places:
[tex]\[ k \approx 0.0608 \][/tex]
3. Predict the number of bacteria after 19 hours:
Using the formula again \(P = A e^{k t}\):
[tex]\[ P_{19} = P_0 e^{k t_{19}} \][/tex]
Substitute \(P_0\), \(k\), and \(t_{19}\):
[tex]\[ P_{19} = 6000 e^{0.0608 \times 19} \][/tex]
Compute the exponent:
[tex]\[ 0.0608 \times 19 = 1.1552 \][/tex]
Then:
[tex]\[ P_{19} = 6000 e^{1.1552} \][/tex]
Calculate \(e^{1.1552}\):
[tex]\[ e^{1.1552} \approx 3.1731 \][/tex]
Finally, calculate \(P_{19}\):
[tex]\[ P_{19} = 6000 \times 3.1731 \approx 19038.4889 \][/tex]
4. Round the answer to the nearest whole number:
[tex]\[ P_{19} \approx 19,038 \][/tex]
Therefore, after 19 hours, there will be approximately 19,038 bacteria present in the culture.
[tex]\[ P = A e^{k t} \][/tex]
1. Initial Information:
- Initial number of bacteria (\(P_0\)) = 6,000
- Number of bacteria after 3 hours (\(P_3\)) = 7,200
- Time elapsed to reach 7,200 bacteria (\(t_3\)) = 3 hours
- Time after which we need to predict the number of bacteria (\(t_{19}\)) = 19 hours
2. Determine the growth rate constant \(k\):
Using the formula \(P = A e^{k t}\):
[tex]\[ P_3 = P_0 e^{k t_3} \][/tex]
Substitute \(P_3\), \(P_0\), and \(t_3\):
[tex]\[ 7200 = 6000 e^{3k} \][/tex]
Solve for \(e^{3k}\):
[tex]\[ e^{3k} = \frac{7200}{6000} \][/tex]
[tex]\[ e^{3k} = 1.2 \][/tex]
Take the natural logarithm on both sides to solve for \(3k\):
[tex]\[ 3k = \ln(1.2) \][/tex]
Divide by 3 to isolate \(k\):
[tex]\[ k = \frac{\ln(1.2)}{3} \][/tex]
Calculate \(k\) and round to four decimal places:
[tex]\[ k \approx 0.0608 \][/tex]
3. Predict the number of bacteria after 19 hours:
Using the formula again \(P = A e^{k t}\):
[tex]\[ P_{19} = P_0 e^{k t_{19}} \][/tex]
Substitute \(P_0\), \(k\), and \(t_{19}\):
[tex]\[ P_{19} = 6000 e^{0.0608 \times 19} \][/tex]
Compute the exponent:
[tex]\[ 0.0608 \times 19 = 1.1552 \][/tex]
Then:
[tex]\[ P_{19} = 6000 e^{1.1552} \][/tex]
Calculate \(e^{1.1552}\):
[tex]\[ e^{1.1552} \approx 3.1731 \][/tex]
Finally, calculate \(P_{19}\):
[tex]\[ P_{19} = 6000 \times 3.1731 \approx 19038.4889 \][/tex]
4. Round the answer to the nearest whole number:
[tex]\[ P_{19} \approx 19,038 \][/tex]
Therefore, after 19 hours, there will be approximately 19,038 bacteria present in the culture.
