Question

the populations ( p ) (in thousands) of a certain city from 2000 through 2007 can be modeled by ( p = 1676.3e^{kt} ), where ( t ) represents the year, with ( t = 0 ) corresponding to 2000. in 2006, the population of the city was about 1,960,000. (a) find the value of ( k ). (round your answer to five decimal places.) ( k=)

Explanation:

Step1: Substitute the values into the formula

Given \(P = 1676.3e^{kt}\), when \(t = 6\) (since \(2006 - 2000=6\)) and \(P = 1960\) (because \(1960000\div1000 = 1960\)).
So, \(1960=1676.3e^{6k}\).

Step2: Solve for \(e^{6k}\)

Divide both sides by \(1676.3\): \(\frac{1960}{1676.3}=e^{6k}\).
\(e^{6k}\approx1.16924\).

Step3: Take the natural logarithm of both sides

\(\ln(e^{6k})=\ln(1.16924)\).
Since \(\ln(e^{x}) = x\), we have \(6k=\ln(1.16924)\).
\(\ln(1.16924)\approx0.15639\).

Step4: Solve for \(k\)

\(k=\frac{\ln(1.16924)}{6}\).
\(k=\frac{0.15639}{6}\approx0.02607\).

Answer:

\(k = 0.02607\)