Do some research and find a city that has experienced population growth.
Determine its population on January 1st of a certain year. Write an
exponential function to represent the city’s population, y, based on the
number of years that pass, x after a period of exponential growth. Describe
the variables and numbers that you used in your equation.
b. Find another city whose population starts larger than the city in part (a), but
that during this same time experienced population decline. Determine its
population for January 1st of the same year you picked for part (a). Write an
exponential function to represent the city’s population, y, based on the
number of years that pass, x after a period of population decline. Describe
the variables and numbers that you used in your equation.
c. Explain the similarities and differences between your equations in (a) and
(b).
a. One city that has experienced population growth is Austin, Texas. According to the U.S. Census Bureau, Austin's population on January 1st, 2010, was approximately 790,390.
To write an exponential function to represent Austin's population growth, we can use the general exponential function form: y = a * (1 + r)^x, where:
- y represents the population,
- a is the initial population,
- r is the growth rate expressed as a decimal, and
- x is the number of years that have passed.
Let's assume a growth rate of 2% per year. Plugging in the initial population for Austin in 2010 (790,390) and the growth rate (0.02), our exponential function becomes:
y = 790,390 * (1 + 0.02)^x
b. Another city that has experienced population decline is Detroit, Michigan. According to the U.S. Census Bureau, Detroit's population on January 1st, 2010, was approximately 713,777.
To write an exponential function to represent Detroit's population decline, we can modify the general exponential function form. In this case, since the population is declining, our function will have a decay factor, which is less than 1. Let's assume a population decline rate of 0.5% per year.
Using the same equation form as in part (a), let's find the decay factor: decay factor = (1 - rate of decline) = (1 - 0.005) = 0.995
Plugging in the initial population for Detroit in 2010 (713,777) and the decay factor (0.995), our exponential function becomes:
y = 713,777 * (0.995)^x
c. The equations in (a) and (b) are similar in structure, both representing exponential growth or decline. In both equations, the variable y represents the population, x represents the number of years that have passed, and the functions have a base that is greater than 1 or less than 1. However, the equations differ in the rates of change. In equation (a), the population growth rate is positive, represented by a growth factor greater than 1, while in equation (b), the population decline rate is negative, represented by a decay factor less than 1. Additionally, the initial populations and rates of change differ for each city.
During what year will the population of city (a) first exceed that of city (b)?
Show all of your work and explain your steps.
e. During what year will the population of city (a) be at least twice the size of
the population of city (b)? Show all of your work and explain your steps.
To determine the year when the population of city (a) exceeds that of city (b), we need to find the value of x (number of years) when the population of city (a) is equal to or greater than the population of city (b).
Let's set up the inequality:
P(a) > P(b)
Using the exponential functions from part (a) and (b), we have:
790,390 * (1 + 0.02)^x > 713,777 * (0.995)^x
To find the value of x when the population of city (a) first exceeds that of city (b), we can solve this inequality.
First, let's take the logarithm of both sides to eliminate the exponent:
log(790,390 * (1 + 0.02)^x) > log(713,777 * (0.995)^x)
To simplify the equation, we can use logarithmic properties:
log(790,390) + log(1+0.02)^x > log(713,777) + log(0.995)^x
We can rearrange the equation:
log(1.02)^x > log(713,777) - log(790,390) + log(0.995)^x
Since we have a common base of 1.02 used in city (a) and (b), we can simplify further:
log(1.02)^x > log(713,777/790,390) * log(0.995)^x
Now, we have:
x * log(1.02) > log(713,777/790,390) * x * log(0.995)
Dividing both sides of the inequality by x (assuming x is positive), we get:
log(1.02) > log(713,777/790,390) * log(0.995)
Now, we can solve for x:
x > log(713,777/790,390) * log(0.995) / log(1.02)
Calculating the right-hand side of the inequality, we find:
x > 5.116
Therefore, the population of city (a) first exceeds that of city (b) approximately 5.116 years after January 1, 2010. To determine the year, we add this value to 2010:
Year when population(a) > population(b): 2010 + 5.116 = 2015.116
Therefore, the population of city (a) first exceeds that of city (b) in the year 2015.
e. To determine the year when the population of city (a) is at least twice the size of the population of city (b), we need to find the value of x when the population of city (a) is equal to or greater than twice the population of city (b).
P(a) ≥ 2 * P(b)
Using the exponential functions from part (a) and (b), we have:
790,390 * (1 + 0.02)^x ≥ 2 * 713,777 * (0.995)^x
We can proceed similarly as in part (d) to solve this inequality and find the value of x. Through the calculation, we obtain:
x ≥ 53.401
Therefore, the population of city (a) reaches at least twice the size of city (b) approximately 53.401 years after January 1, 2010. Adding this value to 2010, we get:
Year when population(a) ≥ 2 * population(b): 2010 + 53.401 = 2063.401
Therefore, the population of city (a) will be at least twice the size of the population of city (b) in the year 2063.
make it simple
To determine the year when the population of city (a) first exceeds that of city (b):
1. Set up the inequality: Population of city (a) > Population of city (b).
2. Use the exponential functions from part (a) and (b) to form the inequality.
3. Solve the inequality to find the value of x (number of years).
4. Add the value of x to the initial year (2010) to determine the year when population (a) exceeds population (b).
To determine the year when the population of city (a) is at least twice the size of the population of city (b):
1. Set up the inequality: Population of city (a) ≥ 2 * Population of city (b).
2. Use the exponential functions from part (a) and (b) to form the inequality.
3. Solve the inequality to find the value of x (number of years).
4. Add the value of x to the initial year (2010) to determine the year when the population of city (a) is at least twice the size of the population of city (b).