Question

the population of a southern city is growing according to the exponential law. complete 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.
let ( n_0 ) represent the initial population of the city and let k be a positive constant that represents the annual growth rate of the citys population.
( n(t)=n_0e^{kt} )
(type an exact answer in terms of e.)

(b) if the population doubled in size over a 13 - month period and the current population is 40,000, what will the population be 5 years from now?
the population will be ( square ) people.
(do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Find the value of \(k\)

The population doubles in \(t = \frac{13}{12}\) years. Using the formula \(N(t)=N_0e^{kt}\), when \(N(t) = 2N_0\), we have \(2N_0=N_0e^{k\times\frac{13}{12}}\).
Divide both sides by \(N_0\) (since \(N_0
eq0\)): \(2 = e^{\frac{13k}{12}}\).
Take the natural logarithm of both sides: \(\ln(2)=\frac{13k}{12}\).
Solve for \(k\): \(k=\frac{12\ln(2)}{13}\).

Step2: Calculate the population after \(t = 5\) years

We know \(N_0 = 40000\) and \(k=\frac{12\ln(2)}{13}\), and use the formula \(N(t)=N_0e^{kt}\).
Substitute \(t = 5\) and \(k=\frac{12\ln(2)}{13}\) into the formula: \(N(5)=40000e^{\frac{12\ln(2)}{13}\times5}\).
Since \(a\ln(b)=\ln(b^{a})\), then \(\frac{12\ln(2)}{13}\times5=\ln(2^{\frac{60}{13}})\).
So \(N(5)=40000\times2^{\frac{60}{13}}\).
Calculate \(2^{\frac{60}{13}}\approx2^{4.615}\approx9.519\).
Then \(N(5)=40000\times9.519 = 380760\).

Answer:

\(380760\)