Answer:
a) The exponential function is [tex]A(t) = 10083(1.068)^t[/tex]
b)
The balance after 1 year is of $10,768.644
The balance after 2 years is of $11,500.91
The balance after 5 years is of $14,010.25.
The balance after 10 years is of $19,467.15
c)
The doubling time is of 10.54 years.
Step-by-step explanation:
Continuously compounded interest:
The amount of money earning after t years, with interest compounded continuously, is given by:
[tex]A(t) = A(0)(1+r)^t[/tex]
In which A(0) is the amount of the initial investment and r is the growth rate, as a decimal.
a) Find the exponential function that describes the amount in the account after time t, in years.
Suppose that $10,083 is invested at an interest rate of 6.8% per year
This means, respectively, that [tex]A(0) = 10083, r = 0.068[/tex]
So
[tex]A(t) = A(0)(1+r)^t[/tex]
[tex]A(t) = 10083(1+0.068)^t[/tex]
[tex]A(t) = 10083(1.068)^t[/tex]
b) What is the balance after 1 year? 2 years? 5 years? 10 years?
After 1 year:
[tex]A(1) = 10083(1.068)^{1} = 10,768.644[/tex]
The balance after 1 year is of $10,768.644
After 2 years:
[tex]A(2) = 10083(1.068)^{2} = 11,500.91[/tex]
The balance after 2 years is of $11,500.91.
After 5 years:
[tex]A(5) = 10083(1.068)^{5} = 14,010.25[/tex]
The balance after 5 years is of $14,010.25.
After 10 years:
[tex]A(10) = 10083(1.068)^{10} = 19,467.15[/tex]
The balance after 10 years is of $19,467.15.
c) What is the doubling time?
This is t for which [tex]A(t) = 2A(0)[/tex]. So
[tex]A(t) = A(0)(1.068)^t[/tex]
[tex](1.068)^t = 2[/tex]
[tex]\log{(1.068)^t} = \log{2}[/tex]
[tex]t\log{1.068} = \log{2}[/tex]
[tex]t = \frac{\log{2}}{\log{1.068}}[/tex]
[tex]t = 10.54[/tex]
The doubling time is of 10.54 years.
