SOLUTION: 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 pop
Algebra.Com
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.
RELATED QUESTIONS
Suppose a city with population 210,000 has been growing at a rate of 3% per year. If this (answered by ikleyn)
In 1980, the population of a city was 6.8 million. By 1992 the population had grown to... (answered by stanbon)
The city of Anville is currently home to 28000 people, and the population has been... (answered by ewatrrr,greenestamps)
The population of a certain city was 680,000 in the year 2000 and is growing at the... (answered by Nate)
A city's population in the year x = 1974 was y = 1,104,100. In 1979 the population was... (answered by ikleyn,ewatrrr,josgarithmetic)
City A's population is 68000, decreasing at a rate of 80 people per year. City B having... (answered by richwmiller,sunitha.pratapani)
Suppose the population of a city 50,000 and is growing 3 % each year.
A. The initial... (answered by dpurcell58)
In 2012 , the population of a city of was 5.75 million. The exponential growth rate... (answered by Alan3354)
Dallastown has a population of 15,300 and is growing at an exponential rate. What will be (answered by fractalier)