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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]

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