How much did you invest at 9% compounded bi-weekly if 10 years later the investment is worth $54,000?

follow this example

http://www.jiskha.com/display.cgi?id=1301189739

I thought it was the same question, but it is not

let the investment be x

i = .06/26 = .0023077
n = 10(26) = 260
x(1.0023077^260) = 54000

solve for x

To calculate the initial investment, we can use the compound interest formula:

A = P(1 + r/n)^(nt)

Where:
A = the future value (in this case, $54,000)
P = the principal investment (the initial investment we want to find)
r = the annual interest rate (9% or 0.09)
n = the number of compounding periods per year (bi-weekly means 26 periods in a year)
t = the number of years (10 years)

By plugging in these values, we can solve for P:

$54,000 = P(1 + 0.09/26)^(26*10)

Now we can solve this equation step by step to find the initial investment:

1. Divide both sides of the equation by (1 + 0.09/26)^(26*10):

$54,000 / (1 + 0.09/26)^(26*10) = P

2. Simplify the exponent:

$54,000 / (1 + 0.09/26)^(260) = P

3. Calculate the value inside the parentheses:

(1 + 0.09/26) = 1.0034615384615385

4. Raise this value to the power of 260:

(1.0034615384615385)^(260) = 1.3121364199262032

5. Divide $54,000 by this result:

$54,000 / 1.3121364199262032 = $41,158.12

Therefore, the initial investment at a 9% interest rate compounded bi-weekly, which would result in $54,000 after 10 years, is approximately $41,158.12.