Question

Fabian has a savings account balance of $1,374.73. The interest rate on the account is 3.9% compounded twice a year. If he opened the account nine years ago, what was the value of his initial deposit?

Answer 1

A=P(1+rn)^nt
1374.73=P(1+0.0392)^2(9)
Solve for P which is the initial amount.

Answer 2

The following formula can be used to get the solution:
\[A=P(1+\frac{r}{n})^{nt}\]
where A is the amount after t years, P is the principal, r is the annual interest rate as a decimal and n is the number of compounding events per year.
Plugging in values we get
\[1374.73=P(1+\frac{0.039}{2})^{9\times 2}\]
Now just solve for P.