Question

The population of a city is modeled by the equation P(t) = 242,700e0.25t where t is measured in years. If the city continues to grow at this rate, (a) What would be the population after 1 year? (b) Set the equation and solve for time to answer: in approximately how many years will it take for the population to reach one million? (Round your answer to two decimal places.)

Answer 1

Answer:(a) To find the population after 1 year, we can simply plug in t = 1 into the given equation and evaluate:

P(1) = 242,700e^(0.25*1) = 278,926.38

Therefore, the population after 1 year would be approximately 278,926.

(b) We want to solve for the time t when the population P(t) reaches one million. In other words, we want to find t such that:

P(t) = 1,000,000

Substituting the given equation for P(t), we get:

242,700e^(0.25t) = 1,000,000

Dividing both sides by 242,700, we get:

e^(0.25t) = 4.120439

Taking the natural logarithm of both sides, we get:

0.25t = ln(4.120439)

Solving for t, we get:

t = (ln(4.120439))/0.25 ≈ 7.77 years

Therefore, it will take approximately 7.77 years (rounded to two decimal places) for the population to reach one million.

Answer 2

(a) To find the population after 1 year, we can simply plug in t = 1 into the given equation and evaluate:

P(1) = 242,700e^(0.25*1) = 278,926.38

Therefore, the population after 1 year would be approximately 278,926.

(b) We want to solve for the time t when the population P(t) reaches one million. In other words, we want to find t such that:

P(t) = 1,000,000

Substituting the given equation for P(t), we get:

242,700e^(0.25t) = 1,000,000

Dividing both sides by 242,700, we get:

e^(0.25t) = 4.120439

Taking the natural logarithm of both sides, we get:

0.25t = ln(4.120439)

Solving for t, we get:

t = (ln(4.120439))/0.25 ≈ 7.77 years

Therefore, it will take approximately 7.77 years (rounded to two decimal places) for the population to reach one million.