This exercise is based on the following table, which lists interest rates on long-term investments (based on 10-year government bonds) in several countries in 2008.
Assuming that you invest $9,000 in the United States (3.9%), how long (to the nearest year) must you wait before your investment is worth $19,000 if the interest is compounded annually?
I got 19 years, its wrong.
Thank you
Pt = Po(1+r)^n.
r = 3.9% / 100% = 0.039 = APR expressed as a decimal.
n = The # of compounding periods.
Pt = 9000(1.039)^n = 19,000.
(1.039)^n = 19000 / 9000 = 2.11111.
Take Log of both sides:
n*Log(1.039) = Log(2.11111).
N = Log(2.11111) / Log(1.039) = 19.53
yrs = 20 yrs.
To solve this problem, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the final amount
P is the initial principal (amount invested)
r is the interest rate per period (expressed as a decimal)
n is the number of compounding periods per year
t is the number of years
In this case, we are given that the initial investment (P) is $9,000, the interest rate (r) is 3.9% (0.039 as a decimal), and we want to find out how long it will take until the investment is worth $19,000 (A). The compounding is done annually (n = 1).
Let's set up the equation and solve for t:
$19,000 = $9,000(1 + 0.039/1)^(1*t)
Dividing both sides of the equation by $9,000 and rearranging, we get:
(19,000 / 9,000) = (1.039)^t
2.1111 ≈ 1.039^t
To solve for t, we can take the logarithm of both sides:
log(2.1111) ≈ log(1.039^t)
Using the logarithmic property log(a^b) = b * log(a), we get:
log(2.1111) ≈ t * log(1.039)
Now you can use a calculator to find the value of t by dividing the logarithm of 2.1111 by the logarithm of 1.039.
After calculating the value of t, round it to the nearest whole number to determine the approximate number of years you must wait for your investment to be worth $19,000.