The population of a southern city follows the exponential law. Use this information to answer parts a and b . (a) If N is the population of the city and t is the time in years, express N as a function of t . \( \mathrm{N}(\mathrm{t})= \) (Type an expression using t as the variable and in terms of \( e \).) (b) If the population doubled in size over 19 months and the current population is 50,000 , what will the population be 3 years from now? The population will be approximately \( \square \) people. (Do not round until the final answer. Then round to the nearest whole number as needed.)

To express the population \( N \) as a function of time \( t \) in years, we can use the exponential growth formula: \( N(t) = N_0 \cdot e^{kt} \) Where: - \( N_0 \) is the initial population, which we'll find later, - \( k \) is the growth constant, - \( t \) is the time in years. Now for part (b), since the population doubles in size over 19 months (which is 19/12 years), we can set up the equation: \( 2N_0 = N_0 \cdot e^{k(19/12)} \) This simplifies to: \( 2 = e^{k(19/12)} \) Taking the natural logarithm of both sides gives us: \( k(19/12) = \ln(2) \) From this, we solve for \( k \): \( k = \frac{12 \ln(2)}{19} \) Now, if the current population is 50,000, then \( N_0 = 50000 \). We can now find the population 3 years from now (which is 3 years + 19/12 years since the population has doubled) using: \( N(3) = 50000 \cdot e^{k \cdot 3} \) Plugging in the value of \( k \): \( N(3) = 50000 \cdot e^{\left(\frac{12 \ln(2)}{19}\right) \cdot 3} \) Now calculating: \( N(3) = 50000 \cdot e^{\left(\frac{36 \ln(2)}{19}\right)} \) Calculating the exponent and then \( N(3) \): Finally, compute this for approximation: 1. Calculate \( e^{\left(\frac{36 \ln(2)}{19}\right)} \) 2. Multiply by 50000 This will yield approximately \( 91841 \) people when we round to the nearest whole number. Hence, the population will be approximately \( 91841 \) people in 3 years.