Posted by **Annie** on Sunday, June 8, 2008 at 3:48pm.

I'm pretty confused about these problems. We're learning growth and decay, but there are quite a few formulas.

1. Suppose $500 is invested at 6% annual interest compounded twice a year. When will the investment be worth $1000?

2. Suppose $500 is invested at 6% annual interest compounded continuously. When will the investment be worth $1000?

... I'm really confused

- Algebra II -
**Damon**, Sunday, June 8, 2008 at 4:20pm
If the annual interest is 6 percent every half year the amount in the account is multiplied by 1.03 every half year.

so after n half years the account will be worth:

$500 * (1.03)* (1.03) * (1.03) etc n times

or

$500 * (1.03)^n

so do the problem in half years

$500 * (1.03)^n = 1000

1.03^n = 2

take the log of both sides remembering that log x^y = y log x

n * .012837 = .301030

n = 23.45 half years = 11.7 years

-----------------------------------

Now for the continuous problem you either have to know the formula from your text book or know calculus.

Since I do not have a textbook, I will figure it out and find the formula

dy/dt = .06 y

dy/y = .06 dt

ln y = .06 t + c

y = e^(.06 t + c) = C e^.06 t

when t = 0, y = $500 so

THIS NEXT LINE IS PROBABLY IN YOUR TEXT

y = 500 e^.06 t

so we want

1000 = 500 e^.06 t

ln 2 = .06 t

.693147 = .06 t

so t = 11.55 years

Just a little better than compounding twice a year

- Algebra II -
**Damon**, Sunday, June 8, 2008 at 4:22pm
By the way, a rule of thumb is that your money doubles in ten years at 7% interest.

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