Question

in 2012, the population of a city was 63,000. by 2017, the population was reduced to approximately 54,100. identify any equations that are appropriate models for the population of the city, and explain why the others are not

Answer 1

`f(x) = 63000(0.97)^x`

Explanation:

Given

`f(0) = 63000` ---x = 0, in 2012

`f(5) = 54100` -- x = 5, in 2017

Required

Select all possible equations

Because there is a reduction in the population, as time increases; the rate must be less than 1.

An exponential function is represented as:

`f(x) = ab^x`

Where

`b = rate`

rate > 1 in options (a) and (b) i.e. 1.03

This implies that (a) and (b) cannot be true

For option (c), we have:

`f(x) = 63000(0.97)^x`

Set x = 0

`f(0) = 63000(0.97)^0 = 63000*1=63000\\`

Set x = 5

`f(5) = 63000(0.97)^5 = 63000*0.8587=54098.1 \approx 54100`

This is true because the calculated values of f(0) and f(5) correspond to the given values

For option (d), we have:

`f(x) = 52477(0.97)^x`

Set x = 0

`f(0) = 52477(0.97)^0 - 52477* 1 = 52477`

This is false because the calculated value of f(0) does not correspond to the given value

For option (e), we have:

`f(x) = 63000(0.97)^(1)/(5x)`

Set x = 0

`f(0) = 63000(0.97)^(1)/(5*0) = 63000(0.97)^(1)/(0) =`undefined

This is false because the f(x) is not undefined at x = 0

For option (f), we have:

`f(x) = 52477(0.97)^{5x`

Set x = 0

`f(0) = 52477(0.97)^(5*0) = 52477(0.97)^0 =52477*1= 52477`

This is false because the calculated value of f(0) does not correspond to the given value

From the computations above, only (c)
`f(x) = 63000(0.97)^x` is true