Using compound interest, it is found that:
a) $1,270.70 will be in the account after 6 years.
b) It will take 17.53 years for the $1000 to double.
What is compound interest?
The amount of money earned, in compound interest, after t years, is given by:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
In which:
- A(t) is the amount of money after t years.
- P is the principal(the initial sum of money).
- r is the interest rate(as a decimal value).
- n is the number of times that interest is compounded per year.
- t is the time in years for which the money is invested or borrowed.
In this problem, the parameters are given as follows:
[tex]A(0) = 1000, r = 0.04, n = 12[/tex]
Item a:
Over 6 years, that is t = 6, thus:
[tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]
[tex]A(t) = 1000\left(1 + \frac{0.04}{12}\right)^{12 \times 6}[/tex]
[tex]A(t) = 1270.7[/tex]
$1,270.70 will be in the account after 6 years.
Item b:
This is t for which A(t) = 2P = 2000, hence:
[tex]2000 = 1000\left(1 + \frac{0.04}{12}\right)^{12t}[/tex]
[tex](1.0033)^{12t} = 2[/tex]
[tex]\log{(1.0033)^{12t}} = \log{2}[/tex]
[tex]12\log{(1.0033)} = \log{2}[/tex]
[tex]t = \frac{\log{2}}{12\log{1.0033}}[/tex]
[tex]t = 17.53[/tex]
It will take 17.53 years for the $1000 to double.
More can be learned about compound interest at https://brainly.com/question/25781328
