Question

Devise the exponential growth function that fits the given data, then answer the accompanying question. Be sure to identify the refernce point (t=0) and units of time.Between 2003 and 2008, the average rate of inflation in a certain country was about 4% per year. If a cart of groceries cost $120 in 2003, what will it cost in 2013 assuming the rate of inflation remains constant?

Answer 1

The cost of groceries in 2013, after applying an annual inflation rate of 4% for 10 years, will be approximately $177.63.

Step-by-step explanation:

To calculate the annual rate of inflation and the cost in 2013, you can use the exponential growth function, where cost =
`initial_(cost) * (1 + rate)^{time`. In this case, the initial cost in 2003 (t=0) is $120 and the annual rate of inflation is 4%, or 0.04. To calculate the cost in 2013, which is 10 years after 2003, apply the exponential growth function:

Cost in 2013 = $120 * (1 + 0.04)¹⁰

This calculation yields:

Cost in 2013 = $120 * (1.04)¹⁰

Cost in 2013 = $120 * 1.48024

Cost in 2013 = $177.63

Therefore, in 2013, assuming the rate of inflation remains constant, the cart of groceries will cost approximately $177.63.

Answer 2

`A(t=10) = 120 e^(ln(1.04)10)=177.629`

And that would be the approximately cost for 2013.

Step-by-step explanation:

For this case we need to define some notation first.

A= population , t= represent the years after 2003, C= constant for the exponential model.

The starting point t=0 correspond to the year of 2003.

On this case we are assuming the following exponential model:

`A(t) = A_o e^(Ct)`

The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:

`120=A_o e^(C(0))=A_o e^0 = A_o`

And then the model is:

`A(t) =120 e^(Ct)`

Now we need to determine the value for C. Since we know that inflation increase 4% per year we have that after one year we have 1.04 times the value of the original value, and we have this equation:

`1.04 A_o= A_o e^(C(1))= A_o e^C`

And we got this:

`1.04= e^C`

Applying ln on both sides we got:

`ln(1.04)= C=0.0392207`

So then our model is given by:

`A(t) = 120 e^(ln(1.04)t)`

For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:

`A(t=10) = 120 e^(ln(1.04)10)=177.629`

And that would be the approximately cost for 2013.