Answer:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
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:
[tex]A(t) = A_o e^{Ct}[/tex]
The initial value on this case is for t=0 A(t=0)= 120 and if we replace we got this:
[tex]120=A_o e^{C(0)}=A_o e^0 = A_o[/tex]
And then the model is:
[tex]A(t) =120 e^{Ct}[/tex]
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:
[tex]1.04 A_o= A_o e^{C(1)}= A_o e^C[/tex]
And we got this:
[tex]1.04= e^C [/tex]
Applying ln on both sides we got:
[tex]ln(1.04)= C=0.0392207[/tex]
So then our model is given by:
[tex]A(t) = 120 e^{ln(1.04)t}[/tex]
For 2013 we have that t=10 since 2013-2003 = 10 after 2003, if we replace t=10 we got this:
[tex]A(t=10) = 120 e^{ln(1.04)10}=177.629[/tex]
And that would be the approximately cost for 2013.
