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
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
By the way, a rule of thumb is that your money doubles in ten years at 7% interest.
I can help you with these problems! Let's break them down one by one.
1. For the first problem, we have an investment of $500 with an annual interest rate of 6% compounded twice a year. We want to find out when the investment will be worth $1000.
To solve this, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = time in years
In this case, P = $500, r = 6% = 0.06 (as a decimal), n = 2 (compounded twice a year), and we need to find t.
So, the formula becomes:
$1000 = $500(1 + 0.06/2)^(2t)
To solve for t, we can rearrange the equation:
(1 + 0.06/2)^(2t) = 2
Now we need to solve for t by taking the logarithm of both sides of the equation. Since the base is 1.03, we can use the natural logarithm (ln) or logarithm base 10 (log):
ln((1 + 0.06/2)^(2t)) = ln(2)
2t * ln(1 + 0.06/2) = ln(2)
t = ln(2) / (2 * ln(1 + 0.06/2))
Calculating this value will give us the time it takes for the investment to be worth $1000.
2. For the second problem, we have the same initial investment of $500 and an interest rate of 6%. However, this time the interest is compounded continuously.
The formula for continuous compound interest is:
A = P * e^(rt)
Where:
A = the future value of the investment
P = the principal amount (initial investment)
r = annual interest rate (as a decimal)
t = time in years
e = Euler's number (approximately 2.71828)
In this case, P = $500, r = 6% = 0.06 (as a decimal), and we need to find t.
So, the formula becomes:
$1000 = $500 * e^(0.06t)
To solve for t, we can rearrange the equation:
e^(0.06t) = 2
Now, we can solve for t using natural logarithms:
ln(e^(0.06t)) = ln(2)
0.06t = ln(2)
t = ln(2) / 0.06
Calculate this value, and you will get the time it takes for the investment to be worth $1000 when compounded continuously.
Remember, these formulas provide a way to calculate future values based on compound interest, and depending on the context of the problem, you can use the appropriate formula.