1. **Stating the problem:** We are given a function for quantity $Q$ as a function of time $t$: $$Q = 9e^{kt}$$ where the constant $k$ is given by: $$k = \frac{\ln(1/2)}{3.7}$$ We need to find the derivative $Q'$ (the rate of change of $Q$ with respect to $t$) and understand the amount at a given time. 2. **Formula for derivative of exponential functions:** If $Q = Ae^{kt}$, then the derivative with respect to $t$ is: $$Q' = Ake^{kt}$$ This comes from the chain rule, where the derivative of $e^{kt}$ is $ke^{kt}$. 3. **Calculate $k$ explicitly:** Given: $$k = \frac{\ln(1/2)}{3.7}$$ Recall that $\ln(1/2) = -\ln(2)$, so: $$k = \frac{-\ln(2)}{3.7}$$ 4. **Write the expression for $Q'$:** Using the formula for the derivative: $$Q' = 9 \times k \times e^{kt} = 9k e^{kt}$$ 5. **Summary:** - The original function is: $$Q = 9e^{kt}$$ - The constant is: $$k = \frac{\ln(1/2)}{3.7}$$ - The derivative (rate of change) is: $$Q' = 9k e^{kt}$$ This derivative tells us how fast the quantity $Q$ changes at any time $t$. **Final answers:** $$k = \frac{\ln(1/2)}{3.7}$$ $$Q' = 9k e^{kt}$$