Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

City A had a population of 10000 in the year 1990. City A’s population grows at a constant rate of 3% per year. City B has a population that is growing exponentially. In the year 2000, there were half as many people in B as in A. In the year 2010, the population of A was 20% more than the population of B.
When will the populations be equal? Give your answer in years after 1990.

Sagot :

MrRoyal

Answer:

City A and city B will have equal population 25years after 1990

Step-by-step explanation:

Given

Let

[tex]t \to[/tex] years after 1990

[tex]A_t \to[/tex] population function of city A

[tex]B_t \to[/tex] population function of city B

City A

[tex]A_0 = 10000[/tex] ---- initial population (1990)

[tex]r_A =3\%[/tex] --- rate

City B

[tex]B_{10} = \frac{1}{2} * A_{10}[/tex] ----- t = 10 in 2000

[tex]A_{20} = B_{20} * (1 + 20\%)[/tex] ---- t = 20 in 2010

Required

When they will have the same population

Both functions follow exponential function.

So, we have:

[tex]A_t = A_0 * (1 + r_A)^t[/tex]

[tex]B_t = B_0 * (1 + r_B)^t[/tex]

Calculate the population of city A in 2000 (t = 10)

[tex]A_t = A_0 * (1 + r_A)^t[/tex]

[tex]A_{10} = 10000 * (1 + 3\%)^{10}[/tex]

[tex]A_{10} = 10000 * (1 + 0.03)^{10}[/tex]

[tex]A_{10} = 10000 * (1.03)^{10}[/tex]

[tex]A_{10} = 13439.16[/tex]

Calculate the population of city A in 2010 (t = 20)

[tex]A_t = A_0 * (1 + r_A)^t[/tex]

[tex]A_{20} = 10000 * (1 + 3\%)^{20}[/tex]

[tex]A_{20} = 10000 * (1 + 0.03)^{20}[/tex]

[tex]A_{20} = 10000 * (1.03)^{20}[/tex]

[tex]A_{20} = 18061.11[/tex]

From the question, we have:

[tex]B_{10} = \frac{1}{2} * A_{10}[/tex]  and  [tex]A_{20} = B_{20} * (1 + 20\%)[/tex]

[tex]B_{10} = \frac{1}{2} * A_{10}[/tex]

[tex]B_{10} = \frac{1}{2} * 13439.16[/tex]

[tex]B_{10} = 6719.58[/tex]

[tex]A_{20} = B_{20} * (1 + 20\%)[/tex]

[tex]18061.11 = B_{20} * (1 + 20\%)[/tex]

[tex]18061.11 = B_{20} * (1 + 0.20)[/tex]

[tex]18061.11 = B_{20} * (1.20)[/tex]

Solve for B20

[tex]B_{20} = \frac{18061.11}{1.20}[/tex]

[tex]B_{20} = 15050.93[/tex]

[tex]B_{10} = 6719.58[/tex] and [tex]B_{20} = 15050.93[/tex] can be used to determine the function of city B

[tex]B_t = B_0 * (1 + r_B)^t[/tex]

For: [tex]B_{10} = 6719.58[/tex]

We have:

[tex]B_{10} = B_0 * (1 + r_B)^{10}[/tex]

[tex]B_0 * (1 + r_B)^{10} = 6719.58[/tex]

For: [tex]B_{20} = 15050.93[/tex]

We have:

[tex]B_{20} = B_0 * (1 + r_B)^{20}[/tex]

[tex]B_0 * (1 + r_B)^{20} = 15050.93[/tex]

Divide [tex]B_0 * (1 + r_B)^{20} = 15050.93[/tex] by [tex]B_0 * (1 + r_B)^{10} = 6719.58[/tex]

[tex]\frac{B_0 * (1 + r_B)^{20}}{B_0 * (1 + r_B)^{10}} = \frac{15050.93}{6719.58}[/tex]

[tex]\frac{(1 + r_B)^{20}}{(1 + r_B)^{10}} = 2.2399[/tex]

Apply law of indices

[tex](1 + r_B)^{20-10} = 2.2399[/tex]

[tex](1 + r_B)^{10} = 2.2399[/tex] --- (1)

Take 10th root of both sides

[tex]1 + r_B = \sqrt[10]{2.2399}[/tex]

[tex]1 + r_B = 1.08[/tex]

Subtract 1 from both sides

[tex]r_B = 0.08[/tex]

To calculate [tex]B_0[/tex], we have:

[tex]B_0 * (1 + r_B)^{10} = 6719.58[/tex]

Recall that: [tex](1 + r_B)^{10} = 2.2399[/tex]

So:

[tex]B_0 * 2.2399 = 6719.58[/tex]

[tex]B_0 = \frac{6719.58}{2.2399}[/tex]

[tex]B_0 = 3000[/tex]

Hence:

[tex]B_t = B_0 * (1 + r_B)^t[/tex]

[tex]B_t = 3000 * (1 + 0.08)^t[/tex]

[tex]B_t = 3000 * (1.08)^t[/tex]

The question requires that we solve for t when:

[tex]A_t = B_t[/tex]

Where:

[tex]A_t = A_0 * (1 + r_A)^t[/tex]

[tex]A_t = 10000 * (1 + 3\%)^t[/tex]

[tex]A_t = 10000 * (1 + 0.03)^t[/tex]

[tex]A_t = 10000 * (1.03)^t[/tex]

and

[tex]B_t = 3000 * (1.08)^t[/tex]

[tex]A_t = B_t[/tex] becomes

[tex]10000 * (1.03)^t = 3000 * (1.08)^t[/tex]

Divide both sides by 10000

[tex](1.03)^t = 0.3 * (1.08)^t[/tex]

Divide both sides by [tex](1.08)^t[/tex]

[tex](\frac{1.03}{1.08})^t = 0.3[/tex]

[tex](0.9537)^t = 0.3[/tex]

Take natural logarithm of both sides

[tex]\ln(0.9537)^t = \ln(0.3)[/tex]

Rewrite as:

[tex]t\cdot\ln(0.9537) = \ln(0.3)[/tex]

Solve for t

[tex]t = \frac{\ln(0.3)}{ln(0.9537)}[/tex]

[tex]t = 25.397[/tex]

Approximate

[tex]t = 25[/tex]

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.