Question

a. in 2000, the population of a country was approximately 6.41 million and by 2060 it is projected to grow to 11 million. use the exponential growth model $a = a_0e^{kt}$, in which t is the number of years after 2000 and $a_0$ is in millions, to find an exponential growth function that models the data. b. by which year will the population be 8 million? a. the exponential growth function that models the data is $a = \square$. (simplify your answer. use integers or decimals for any numbers in the expression. round to two decimal places as needed.) b. the countrys population will be 8 million in the year $\square$. (round to the nearest year as needed.)

Explanation:

Step1: Find the value of \(k\)

Given \(A = A_0e^{kt}\), when \(t = 0\) (year 2000), \(A_0=6.41\). When \(t = 60\) (year 2060), \(A = 11\).
Substitute into the formula: \(11=6.41e^{60k}\).
First, divide both sides by \(6.41\): \(\frac{11}{6.41}=e^{60k}\).
Take the natural logarithm of both sides: \(\ln(\frac{11}{6.41})=\ln(e^{60k})\).
Since \(\ln(e^{x})=x\), we have \(60k=\ln(\frac{11}{6.41})\).
Calculate \(\ln(\frac{11}{6.41})\approx\ln(1.716)=0.541\).
Then \(k=\frac{0.541}{60}\approx0.009\).

Step2: Write the exponential growth function

Substitute \(A_0 = 6.41\) and \(k\approx0.009\) into \(A = A_0e^{kt}\), we get \(A = 6.41e^{0.009t}\).

Step3: Solve for \(t\) when \(A = 8\)

Substitute \(A = 8\) into \(A = 6.41e^{0.009t}\): \(8=6.41e^{0.009t}\).
Divide both sides by \(6.41\): \(\frac{8}{6.41}=e^{0.009t}\).
Take the natural logarithm of both sides: \(\ln(\frac{8}{6.41})=\ln(e^{0.009t})\).
Since \(\ln(e^{x})=x\), \(0.009t=\ln(\frac{8}{6.41})\).
Calculate \(\ln(\frac{8}{6.41})\approx\ln(1.248)=0.221\).
Then \(t=\frac{0.221}{0.009}\approx24.56\).
Since \(t\) is the number of years after 2000, the year is \(2000 + 25=2025\) (rounded to the nearest year).

Answer:

a. \(A = 6.41e^{0.009t}\)
b. \(2025\)