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Step-by-step Solution
Step 1: Express the growth of bacteria as a differential equation
The rate of change of the bacteria population is proportional to the number of bacteria present. Mathematically, this is written as:
$ \frac{dx}{dt} = \lambda x $
where $x$ is the number of bacteria and $\lambda$ is a constant of proportionality.
Step 2: Solve the differential equation
We separate the variables and integrate:
$ \int \frac{dx}{x} = \int \lambda \, dt $
Upon integrating, we get:
$ \ln x = \lambda t + C $
where $C$ is the constant of integration. Exponentiating both sides gives the general solution:
$ x = e^{\lambda t + C} = e^C \cdot e^{\lambda t}. $
Let $e^C = x_0$. Since the bacteria count at $t = 0$ is 1000, we have $x_0 = 1000$. Therefore,
$ x(t) = 1000 \, e^{\lambda t}. $
Step 3: Use initial condition to find the constant $\lambda$
It is given that the bacteria count increases by 20% in 2 hours, i.e., from 1000 to 1200. Therefore, at $t = 2$:
$ x(2) = 1200 = 1000 \, e^{\lambda \cdot 2}. $
Dividing both sides by 1000, we get:
$ 1.2 = e^{2 \lambda} \quad \Longrightarrow \quad 2 \lambda = \ln(1.2) = \ln\left(\frac{6}{5}\right).
$
Hence,
$ \lambda = \frac{1}{2} \ln\left(\frac{6}{5}\right). $
Step 4: Write down the general form with the found $\lambda$
The expression for the population at any time $t$ becomes:
$ x(t) = 1000 \, e^{\left(\frac{t}{2}\right)\ln\left(\frac{6}{5}\right)}.
$
Step 5: Apply the given time condition for $x(t) = 2000$
We are told that $x = 2000$ at
$ t = \frac{k}{\ln\left(\frac{6}{5}\right)}. $
Substituting into the expression, we get:
$ 2000 = 1000 \, e^{\left(\frac{k}{\ln\left(\frac{6}{5}\right)}\right)\left(\frac{\ln\left(\frac{6}{5}\right)}{2}\right)}. $
Simplifying inside the exponent:
$ 2000 = 1000 \, e^{\frac{k}{2}} \quad \Longrightarrow \quad 2 = e^{\frac{k}{2}}.
$
Step 6: Solve for $k$
Taking the natural logarithm on both sides gives:
$ \ln(2) = \frac{k}{2} \quad \Longrightarrow \quad k = 2 \ln(2).
Therefore,
$ \frac{k}{\ln(2)} = 2.
Step 7: Compute the required quantity
We want to find the value of
$ \left(\frac{k}{\ln(2)}\right)^2. $
Since $ \frac{k}{\ln(2)} = 2, $ we have:
$ \left(\frac{k}{\ln(2)}\right)^2 = 2^2 = 4.
Final Answer
$ \left(\frac{k}{\ln(2)}\right)^2 = 4.
