SOLUTION: Suppose $500 is invested at 6% annual interest comounded twice a year. When will the investment be worth $1000? And how do you find this answer?
Question 429573: Suppose $500 is invested at 6% annual interest comounded twice a year. When will the investment be worth $1000? And how do you find this answer? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Suppose $500 is invested at 6% annual interest comounded twice a year. When will the investment be worth $1000? And how do you find this answer?
------
Formula: A(t) = P(1+(r/n))^(nt)
where A(t) is the value after t years.
r is the annual interest rate.
n is the # of compoundings per year
t is the # of years.
-----------------------
1000 = 500(1+(0.06/2))^(2t)
2 = (1.03)^(2t)
------
Take the log and solve for "t":
(2t)*log(1.03) = log(2)
---
2t = log(2)/log(1.03)
---
2t = 23.45
t = 16.72
----
Value will be 1000 sometime during the 16th year.
Cheers,
Stan H.
=========
(