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

Answer:

$A(t=10) = 120 e^{ln(1.04)10}=177.629$

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.