Answer :
To solve this problem, we will use the exponential growth formula:
[tex]\[ P = A e^{kt} \][/tex]
Where:
- \( P \) is the number of bacteria at time \( t \).
- \( A \) is the initial number of bacteria.
- \( k \) is the growth rate constant.
- \( t \) is the time in hours.
Given:
- The initial number of bacteria, \( A \), is 2,000.
- After 2 hours, the number of bacteria, \( P \), is 2,400.
- We want to predict the number of bacteria after 10 hours.
First, we need to determine the growth rate constant \( k \).
1. Substitute the known values into the formula when \( t = 2 \) hours:
[tex]\[ 2400 = 2000 e^{2k} \][/tex]
2. Divide both sides by 2000:
[tex]\[ \frac{2400}{2000} = e^{2k} \][/tex]
[tex]\[ 1.2 = e^{2k} \][/tex]
3. Take the natural logarithm of both sides to solve for \( 2k \):
[tex]\[ \ln(1.2) = 2k \][/tex]
4. Calculate \( \ln(1.2) \):
[tex]\[ \ln(1.2) \approx 0.1823 \][/tex]
5. Solve for \( k \) by dividing by 2:
[tex]\[ k = \frac{0.1823}{2} \][/tex]
[tex]\[ k \approx 0.09115 \][/tex]
Next, we use the value of \( k \) to predict the number of bacteria after 10 hours.
1. Substitute the values \( A = 2000 \), \( k \approx 0.09115 \), and \( t = 10 \) into the formula:
[tex]\[ P = 2000 e^{0.09115 \times 10} \][/tex]
2. Multiply \( k \) by 10:
[tex]\[ 0.09115 \times 10 \approx 0.9115 \][/tex]
3. Calculate \( e^{0.9115} \):
[tex]\[ e^{0.9115} \approx 2.4885 \][/tex]
4. Multiply this result by the initial number of bacteria, 2000:
[tex]\[ P \approx 2000 \times 2.4885 \][/tex]
[tex]\[ P \approx 4977 \][/tex]
Therefore, the predicted number of bacteria present after 10 hours is approximately 4977.
[tex]\[ P = A e^{kt} \][/tex]
Where:
- \( P \) is the number of bacteria at time \( t \).
- \( A \) is the initial number of bacteria.
- \( k \) is the growth rate constant.
- \( t \) is the time in hours.
Given:
- The initial number of bacteria, \( A \), is 2,000.
- After 2 hours, the number of bacteria, \( P \), is 2,400.
- We want to predict the number of bacteria after 10 hours.
First, we need to determine the growth rate constant \( k \).
1. Substitute the known values into the formula when \( t = 2 \) hours:
[tex]\[ 2400 = 2000 e^{2k} \][/tex]
2. Divide both sides by 2000:
[tex]\[ \frac{2400}{2000} = e^{2k} \][/tex]
[tex]\[ 1.2 = e^{2k} \][/tex]
3. Take the natural logarithm of both sides to solve for \( 2k \):
[tex]\[ \ln(1.2) = 2k \][/tex]
4. Calculate \( \ln(1.2) \):
[tex]\[ \ln(1.2) \approx 0.1823 \][/tex]
5. Solve for \( k \) by dividing by 2:
[tex]\[ k = \frac{0.1823}{2} \][/tex]
[tex]\[ k \approx 0.09115 \][/tex]
Next, we use the value of \( k \) to predict the number of bacteria after 10 hours.
1. Substitute the values \( A = 2000 \), \( k \approx 0.09115 \), and \( t = 10 \) into the formula:
[tex]\[ P = 2000 e^{0.09115 \times 10} \][/tex]
2. Multiply \( k \) by 10:
[tex]\[ 0.09115 \times 10 \approx 0.9115 \][/tex]
3. Calculate \( e^{0.9115} \):
[tex]\[ e^{0.9115} \approx 2.4885 \][/tex]
4. Multiply this result by the initial number of bacteria, 2000:
[tex]\[ P \approx 2000 \times 2.4885 \][/tex]
[tex]\[ P \approx 4977 \][/tex]
Therefore, the predicted number of bacteria present after 10 hours is approximately 4977.
