Growth of bacteria is directly proportional to number of bacteria . At t=0 , number of bacteria = 1000 and after 2 hours population is increased by `20%`. After this population becomes 2000 where `t=k/(ln(6/5)`. Find value of `(k/ln2)^2`

Options

A. 4
B. 6
C. 16
D. 8

Answer

To solve the problem step by step, we will follow the mathematical principles of exponential growth and integration.

### Step 1: Set up the differential equation
The growth of bacteria is directly proportional to the number of bacteria, which can be expressed mathematically as:
\[ \frac{dp}{dt} = ap \]
where \( p \) is the population of bacteria and \( a \) is a constant of proportionality.

### Step 2: Integrate the equation
We can separate the variables and integrate:
\[ \frac{dp}{p} = a \, dt \]
Integrating both sides gives:
\[ \ln p = at + C \]
where \( C \) is the constant of integration.

### Step 3: Solve for the constant using initial conditions
At \( t = 0 \), the number of bacteria \( p = 1000 \):
\[ \ln 1000 = a(0) + C \implies C = \ln 1000 \]
Thus, the equation becomes:
\[ \ln p = at + \ln 1000 \]
Exponentiating both sides:
\[ p = 1000 e^{at} \]

### Step 4: Use the first condition to find \( a \)
After 2 hours, the population increases by 20%:
\[ p(2) = 1000 + 0.2 \times 1000 = 1200 \]
Substituting into the equation:
\[ 1200 = 1000 e^{2a} \]
Dividing both sides by 1000:
\[ 1.2 = e^{2a} \]
Taking the natural logarithm:
\[ \ln(1.2) = 2a \implies a = \frac{\ln(1.2)}{2} \]

### Step 5: Use the second condition to find \( k \)
We know that the population becomes 2000 at a time \( t = \frac{k}{\ln(6/5)} \):
\[ 2000 = 1000 e^{at} \]
Dividing both sides by 1000:
\[ 2 = e^{at} \]
Taking the natural logarithm:
\[ \ln(2) = at \]
Substituting \( t = \frac{k}{\ln(6/5)} \):
\[ \ln(2) = a \cdot \frac{k}{\ln(6/5)} \]
Substituting the value of \( a \):
\[ \ln(2) = \frac{\ln(1.2)}{2} \cdot \frac{k}{\ln(6/5)} \]
Rearranging gives:
\[ k = \frac{2 \ln(2) \ln(6/5)}{\ln(1.2)} \]

### Step 6: Find \( \frac{k}{\ln(2)} \)
Now we need to find \( \left(\frac{k}{\ln(2)}\right)^2 \):
\[ \frac{k}{\ln(2)} = \frac{2 \ln(6/5)}{\ln(1.2)} \]
Thus,
\[ \left(\frac{k}{\ln(2)}\right)^2 = \left(\frac{2 \ln(6/5)}{\ln(1.2)}\right)^2 \]

### Step 7: Calculate the final value
We know that \( \ln(6/5) \) and \( \ln(1.2) \) are related:
\[ \ln(1.2) = \ln\left(\frac{6}{5}\right) \implies \frac{\ln(6/5)}{\ln(1.2)} = 1 \]
Thus,
\[ \left(\frac{k}{\ln(2)}\right)^2 = (2)^2 = 4 \]

### Final Answer
The value of \( \left(\frac{k}{\ln(2)}\right)^2 \) is \( 4 \).

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