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?
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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.
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1000 = 500(1+(0.06/2))^(2t)
2 = (1.03)^(2t)
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Take the log and solve for "t":
(2t)*log(1.03) = log(2)
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2t = log(2)/log(1.03)
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2t = 23.45
t = 16.72
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Value will be 1000 sometime during the 16th year.
Cheers,
Stan H.
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