\( P=500 e^{k t} \) is the general solution of population at time, \( t \). Find the particular solution when the population, \( P \) has increased to 5000 after 2 hour. A \( \quad P=50 e^{1.151 t} \) B \( \quad P=500 e^{1.151 t} \) C \( \quad P=500 e^{2.303 t} \) D \( \quad P=50 e^{2.303 t} \)

To find the particular solution given the population increases to 5000 after 2 hours, we start with the general solution \( P = 500 e^{kt} \). We know \( P = 5000 \) when \( t = 2 \). Substituting these values into the equation, we get: \[ 5000 = 500 e^{2k} \] Dividing both sides by 500 gives us: \[ 10 = e^{2k} \] Taking the natural logarithm of both sides leads to: \[ \ln(10) = 2k \] So we find: \[ k = \frac{\ln(10)}{2} \] Now substituting this value of \( k \) back into the general solution yields: \[ P = 500 e^{\frac{\ln(10)}{2} t} = 500 e^{\ln(10^{0.5}) t} = 500 e^{\ln(10^{t/2})} = 500(10^{t/2}) = 50 e^{\ln(10)t} = 50 e^{2.303 t} \] The correct answer corresponds to option D: \( P=50 e^{2.303 t} \). Further Reading: If you're intrigued by how population growth models are derived or wish to explore more real-world factors affecting populations like migration and birth rates, consider looking into books on demography or ecological studies! They often provide fascinating insights into how such models apply in various scenarios. Real-World Application: Population models are extensively used in urban planning, conservation efforts, and epidemiology. Understanding how populations grow can guide resource allocation, help manage wildlife conservation efforts, or even predict the spread of diseases, making this mathematical approach incredibly practical and crucial in today's world!