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The population of a town grows exponentially. After 1 year, the population is 34,560. After 2 years, the population is 37,325. Provide an equation that can be used to predict, y, the number of people living in the town after x years. (Round population values to the nearest whole number.)

Verified Answer :

carlosego


For this case we have a function of the form:
 \(y = A * (b) ^ x\)
 Where,
 A: initial amount
 b: growth rate
 x: time in years
 We must find the values of A and b.
 For this, we use the following data:
 After 1 year, the population is 34,560:
 \(A * (b) ^ 1 = 34560\)
 After 2 years, the population is 37,325:
 \(A * (b) ^ 2 = 37325\)
 We have a system of two equations with two unknowns.
 By dividing both equations we obtain the value of b:
 \(\frac{A * (b) ^ 2}{A * (b) ^ 1} = \frac{37325}{34560}\)
 \(b = \frac{37325}{34560}\)
 \(b = 1.08\)
 Then, the value of A is obtained by substituting b in any of the equations.
 We have then:
 \(A * (b) ^ 1 = 34560 A * (1.08) ^ 1 = 34560 A * (1.08)= 34560\)
 \(A = \frac{34560}{1.08}\)
 \(A = 32000\)
 Answer:
 an equation that can be used to predict, y, the number of people living in the town after x years is:
 \(y=32000*(1.08)^x\)

Answer:

Equation will be \(y=(32083).(1.08)^{x}\)

Step-by-step explanation:

Since population of a town grows exponentially so the function to represent the population growth will be

\(a_{n}=a_{0}(r)^{n}\)

n = number of years

where \(a_{n}\) = population after n years

\(a_{0}\) = population in the base year or population at 0 year

r = common ratio or the growth with which population is increasing year by year

Now It is given in the question after 1 year, the population is 34560

So the expression will be \(34560=a_{0}(r)^{1}=a_{0}.(r)\)------(1)

again question states that after 2 years, the population is 37325

The expression becomes \(a_{2}=37325=a_{0}.(r)^{2}\)------(2)

Now we divide equation 2 from equation 1

\(\frac{37325}{34560}=\frac{a_{0}.r^{2}}{a_{0}.r}\)

r = 1.08

Now we put the value of "r" in equation 1 to calculate the value of \(a_{0}\)

\(34650=a_{0}.(1.08)\)

\(a_{0}= \frac{34650}{1.08}=32083\)

If y be the number of people living in the town after x years then the function formed will be

\(y=(32083).(1.08)^{x}\)

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