Math Problem

Question

$3600 is invested in an account which compounds interest semiannually. What interest rate (to the nearest 0.01%) is needed so that the balance in the account at the end of 3 years is $4000. 

 

originally I used the formula A=P(1+r/n)^(n)(t)

 

Plugging in the values I was given and 4000 in for the value A but when I plugged my answer back into the equation the balance was not $4000.

Answer 1

4000=3600(1+r/2)^(2*3)   Substitute (semiannually is twice a year n=2)

4000/3600= (1 + r/2)^6     Divide both sides by 3600

10/9 = (1 + r/2)^6             Reduce fraction

1.0177 = (1 + r/2)              Take the sixth root using a calculator

0.0177 = r/2                       Subtract 1

.0354 = r                            Multiply by 2

3.54%                                Multiply by 100 to change to a percent