Answer :
Sure! Let's walk through a detailed, step-by-step solution for predicting the number of bacteria present after 16 hours.
### Step-by-Step Solution:
1. Identify the given information:
- Initial number of bacteria, \( P_0 \): 2000
- Number of bacteria after 6 hours, \( P_6 \): 2400
- Time interval for this growth, \( t \): 6 hours
2. Recall the exponential growth formula:
[tex]\[ P = P_0 e^{kt} \][/tex]
where:
- \( P \) is the population at time \( t \)
- \( P_0 \) is the initial population
- \( k \) is the growth rate constant
- \( t \) is the time
3. First, solve for the growth rate constant \( k \):
- At \( t = 6 \), the population \( P_6 \) is 2400.
- So, the equation becomes:
[tex]\[ 2400 = 2000 e^{6k} \][/tex]
- Divide both sides by 2000:
[tex]\[ \frac{2400}{2000} = e^{6k} \][/tex]
[tex]\[ 1.2 = e^{6k} \][/tex]
- Take the natural logarithm of both sides to solve for \( k \):
[tex]\[ \ln(1.2) = 6k \][/tex]
- So:
[tex]\[ k = \frac{\ln(1.2)}{6} \][/tex]
- Using a calculator:
[tex]\[ k \approx \frac{0.1823}{6} \approx 0.0304 \][/tex]
(rounded to four decimal places)
4. Predict the number of bacteria after 16 hours:
- Now we use the growth formula again with \( t = 16 \):
[tex]\[ P_{16} = 2000 e^{k \cdot 16} \][/tex]
- Substitute \( k \approx 0.0304 \):
[tex]\[ P_{16} = 2000 e^{0.0304 \cdot 16} \][/tex]
- Simplify the exponent:
[tex]\[ P_{16} = 2000 e^{0.4864} \][/tex]
- Using a calculator:
[tex]\[ e^{0.4864} \approx 1.6261 \][/tex]
- Thus:
[tex]\[ P_{16} \approx 2000 \times 1.6261 \approx 3252.2 \][/tex]
5. Round the result:
- The bacteria count after 16 hours, rounded to the nearest whole number, is:
[tex]\[ \boxed{3252} \][/tex]
Hence, after 16 hours, there will be approximately 3252 bacteria.
### Step-by-Step Solution:
1. Identify the given information:
- Initial number of bacteria, \( P_0 \): 2000
- Number of bacteria after 6 hours, \( P_6 \): 2400
- Time interval for this growth, \( t \): 6 hours
2. Recall the exponential growth formula:
[tex]\[ P = P_0 e^{kt} \][/tex]
where:
- \( P \) is the population at time \( t \)
- \( P_0 \) is the initial population
- \( k \) is the growth rate constant
- \( t \) is the time
3. First, solve for the growth rate constant \( k \):
- At \( t = 6 \), the population \( P_6 \) is 2400.
- So, the equation becomes:
[tex]\[ 2400 = 2000 e^{6k} \][/tex]
- Divide both sides by 2000:
[tex]\[ \frac{2400}{2000} = e^{6k} \][/tex]
[tex]\[ 1.2 = e^{6k} \][/tex]
- Take the natural logarithm of both sides to solve for \( k \):
[tex]\[ \ln(1.2) = 6k \][/tex]
- So:
[tex]\[ k = \frac{\ln(1.2)}{6} \][/tex]
- Using a calculator:
[tex]\[ k \approx \frac{0.1823}{6} \approx 0.0304 \][/tex]
(rounded to four decimal places)
4. Predict the number of bacteria after 16 hours:
- Now we use the growth formula again with \( t = 16 \):
[tex]\[ P_{16} = 2000 e^{k \cdot 16} \][/tex]
- Substitute \( k \approx 0.0304 \):
[tex]\[ P_{16} = 2000 e^{0.0304 \cdot 16} \][/tex]
- Simplify the exponent:
[tex]\[ P_{16} = 2000 e^{0.4864} \][/tex]
- Using a calculator:
[tex]\[ e^{0.4864} \approx 1.6261 \][/tex]
- Thus:
[tex]\[ P_{16} \approx 2000 \times 1.6261 \approx 3252.2 \][/tex]
5. Round the result:
- The bacteria count after 16 hours, rounded to the nearest whole number, is:
[tex]\[ \boxed{3252} \][/tex]
Hence, after 16 hours, there will be approximately 3252 bacteria.
