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 . \( \mathrm{N}(\mathrm{t})=\mathrm{N}_{0} e^{\mathrm{kt}} \) (Type an expression using t as the variable and in terms of \( e \).) (b) If the population doubled in size over 11 months and the current population is 90,000 , what will the population be 2 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 \( t \), we start with the general form given for exponential growth: \[ N(t) = N_0 e^{kt} \] Here, \( N_0 \) is the initial population and \( k \) is the growth rate. To determine \( N_0 \), we use the current population of 90,000 at \( t = 0 \): \[ N(0) = 90000 \implies N_0 = 90000 \] Now we need to determine \( k \). Since the population doubles in size over 11 months (approximately \( \frac{11}{12} \) years), we can set up the equation: \[ 2N_0 = N_0 e^{k \cdot \frac{11}{12}} \] Dividing both sides by \( N_0 \): \[ 2 = e^{k \cdot \frac{11}{12}} \] Taking the natural logarithm of both sides: \[ \ln(2) = k \cdot \frac{11}{12} \implies k = \frac{12 \ln(2)}{11} \] Now we can express \( N(t) \): \[ N(t) = 90000 e^{\frac{12 \ln(2)}{11}t} \] Moving on to part (b), we want to find \( N(2) \) (2 years from now). Plugging \( t = 2 \) into the equation: \[ N(2) = 90000 e^{\frac{12 \ln(2)}{11} \cdot 2} = 90000 e^{\frac{24 \ln(2)}{11}} \] This simplifies to: \[ N(2) = 90000 \cdot 2^{\frac{24}{11}} \] Calculating \( 2^{\frac{24}{11}} \) gives approximately \( 4.660 \). Therefore: \[ N(2) \approx 90000 \cdot 4.660 \approx 418800 \] Thus, the population will be approximately \( 418800 \) people.