<PRE><B>Question 1167428</B>
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&#39;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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