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.)
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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.