# Algebra II

posted by on .

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 - ,

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
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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 - ,

By the way, a rule of thumb is that your money doubles in ten years at 7% interest.