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Question 1167428: City A is growing at a rate of 6.5%/year and starts with a population of 100,000. City B has a growth rate of 4.58%/year and starts with a population of 250,000.
(a) Model the population of City A.
(b) Model the population of City B.
(c) What will the population of City A be in 10 years? Estimate using two decimal places.
(d) When will the population of City B reach 300,000? Estimate using two decimal places.
(e)When will the population of City A equal the population of City B? Estimate using
two decimal places.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula to use if f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the growth rate per year = percent growth rate per year / 100
n is the number of years.
for city A, the formula becomes f = 100,000 * (1 + .065) ^ n
for city B, the formula becomes f = 250,000 * (1 + .0458) ^ n
the population for city A in 10 years would be modeled as:
f = 100,000 * (1 + .065) ^ 10
solve for f to get f = 187,713.7465.
the population for city B reaching 300,000 would be modeled as:
300,000 = 250,000 * (1 + .0458) ^ n
divide both sides of this formula by 250,000 to get:
300,000 / 250,000 = (1 + .0458) ^ n
simplify a little and take the log of both sides of this equation to get:
log(300/250) = log((1 + .0458) ^ n)
since log((1 + .0458) ^ n) = n * log(1 + .0458), then the equation becomes:
log(300/250) = n * log(1 + .0458)
solve for n to get:
n = log(300/250) / log(1.0458) = 4.071300423 years.
confirm by replacing n with that in the original equation to get:
300,000 = 250,000 * (1 + .0458) ^ n becomes:
300,000 = 250,000 * (1 + .0458) ^ 4.071300423 which becomes:
300,000 = 300,000
this confirms that the solution is that a population of 250,000 will become 300,000 in 4.071300423 years when the growth rate if 4.58% per year.
the population of city A will become equal to the population of city B when 100,000 * (1 + .065) ^ n is equal to 250,000 * (1 + .0458) ^ n
this is because f1 = 100,000 * (1 + .065) ^ n for city A and f2 = 250,000 * (1 + .0458) ^ n for city B.
f1 is the future value for city A.
f2 is the future value for city B.
for f1 to be equal to f2, then 100,000 * (1 + .065) ^ n must be equal to 250,000 * (1 + .0458) ^ n
the formula to solve is therefore:
100,000 * (1 + .065) ^ n = 250,000 * (1 + .0458) ^ n
divide both sides of this equation by 250,000 and divide both sides of this equation by (1 + .065) ^ n to get:
100,000 / 250,000 = (1 + .0458) ^ n / (1 + .065) ^ n
simplify a little to get:
100/250 = 1.0458 ^ n / 1.065 ^ n
since 1.0458 ^ n / 1.065 ^ n is equal to (1.0458 / 1.065) ^ n, then the formula becomes:
100 / 250 = (1.0458 / 1.065) ^ n
take the log of both sides of this equation to get:
log (100 / 250) = n * log(1.0458 / 1.065)
solve for n to get:
n = log(100 / 250) / log(1.0458 / 1.065)
this gets you:
n = 50.36596703 years.
the formula for city A becomes f = 100,000 * 1.065 ^ 50.36596703.
the formula for city B becomes f = 250,000 * 1.0458 ^ 50.36596703.
solve for f in both formulas to get:
the population for city A will become 2,385,005.821 in 50.36596703 years.
the population for city B will become 2,385,005.821 in 50.36596703 years.
they're the same in 50.36596703 years.
answer to your questions are:
(a) Model the population of City A.
f = 100,000 * 1.065 ^ n
(b) Model the population of City B.
f = 250,000 * 1.0458 ^ n
(c) What will the population of City A be in 10 years? Estimate using two decimal places.
population of city A will be 187,713.75 in 10 years.
(d) When will the population of City B reach 300,000? Estimate using two decimal places.
the population of city B will reach 300,000 in 4.07 years.
(e)When will the population of City A equal the population of City B?
the population of city A will be equal to the population of city B in 50.37 years.
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