Key Concepts and Formulas

• Exponential Growth/Decay: $\frac{dx}{dt} = \lambda x \implies x(t) = x_0 e^{\lambda t}$, where $x(t)$ is the quantity at time $t$, $x_0$ is the initial quantity, and $\lambda$ is the growth/decay constant.
• Logarithm Properties: $a \log_b c = \log_b c^a$, $\log_b x = y \iff x = b^y$, $e^{\ln x} = x$.
• Definite Integrals: $\int_a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Step-by-Step Solution

Step 1: Formulate and Solve the Differential Equation

The problem states that the rate of growth of bacteria is proportional to the number of bacteria present. This can be written as a differential equation: $\frac{dx}{dt} = \lambda x$ where $x(t)$ is the number of bacteria at time $t$, and $\lambda$ is the constant of proportionality (growth rate). We solve it by separation of variables: $\frac{dx}{x} = \lambda dt$ Now, integrate both sides from initial conditions ($t=0, x=x_0$) to a general state ($t, x(t)$): $\int_{x_0}^{x(t)} \frac{dx}{x} = \int_0^t \lambda dt$ $\left[ \ln |x| \right]_{x_0}^{x(t)} = \left[ \lambda t \right]_0^t$ $\ln(x(t)) - \ln(x_0) = \lambda t$ $\ln\left(\frac{x(t)}{x_0}\right) = \lambda t$ $x(t) = x_0 e^{\lambda t}$ Since $x_0 = 1000$, we have $x(t) = 1000 e^{\lambda t}$

Step 2: Determine the Growth Rate Constant ($\lambda$)

The problem states that the number of bacteria increases by 20% in 2 hours. This means that at $t=2$, $x(2) = 1000 + 0.20(1000) = 1200$. Substitute these values into the equation: $1200 = 1000 e^{2\lambda}$ $\frac{1200}{1000} = e^{2\lambda}$ $\frac{6}{5} = e^{2\lambda}$ Taking the natural logarithm of both sides: $\ln\left(\frac{6}{5}\right) = 2\lambda$ $\lambda = \frac{1}{2} \ln\left(\frac{6}{5}\right)$

Step 3: Establish the Specific Growth Model

Substitute the value of $\lambda$ into the equation $x(t) = 1000 e^{\lambda t}$: $x(t) = 1000 e^{\frac{1}{2} \ln\left(\frac{6}{5}\right) t}$ $x(t) = 1000 \left(e^{\ln\left(\frac{6}{5}\right)}\right)^{\frac{t}{2}}$ $x(t) = 1000 \left(\frac{6}{5}\right)^{\frac{t}{2}}$

Step 4: Finding the Value of $k$

The problem states that the population of bacteria is 2000 after $t = \frac{k}{\ln\left(\frac{6}{5}\right)}$ hours. Substitute $x(t) = 2000$ and $t = \frac{k}{\ln\left(\frac{6}{5}\right)}$ into the equation: $2000 = 1000 \left(\frac{6}{5}\right)^{\frac{1}{2} \frac{k}{\ln\left(\frac{6}{5}\right)}}$ $2 = \left(\frac{6}{5}\right)^{\frac{k}{2\ln\left(\frac{6}{5}\right)}}$ Take the natural logarithm of both sides: $\ln 2 = \ln\left[ \left(\frac{6}{5}\right)^{\frac{k}{2\ln\left(\frac{6}{5}\right)}} \right]$ $\ln 2 = \frac{k}{2\ln\left(\frac{6}{5}\right)} \ln\left(\frac{6}{5}\right)$ $\ln 2 = \frac{k}{2}$ $k = 2 \ln 2$

Step 5: Calculating the Final Expression

We are asked to find the value of $\left(\frac{k}{\ln 2}\right)^2$. We have $k = 2 \ln 2$, so $\frac{k}{\ln 2} = \frac{2 \ln 2}{\ln 2} = 2$ Therefore, $\left(\frac{k}{\ln 2}\right)^2 = (2)^2 = 4$

Common Mistakes & Tips

• Carefully apply logarithm and exponential properties to simplify expressions.
• Double-check calculations, especially when dealing with fractions and exponents.
• Remember that $\ln x$ is the same as $\log_e x$.

Summary

We started by formulating and solving the differential equation for exponential growth. Then, we used the given information about the bacteria growth to find the growth rate constant, $\lambda$. We then substituted this value back into the equation and used the second condition (population of 2000) to solve for $k$. Finally, we calculated the value of $\left(\frac{k}{\ln 2}\right)^2$, which equals 4.

The final answer is \boxed{4}, which corresponds to option (D).