How much should be invested today at 6.15% compounded monthly to have $12,000 in 7 years?
12,000=P(1+.0615/12)^(7*12)
take ln of each side...
ln(12000)=lnP + 7*12*Ln(1+.0615/12)
lnP=ln(12000)-7*12*ln(1+.0615/12)
pasting this into the google search window:
ln(12000)-7*12*ln(1+.0615/12)=
8.963
so P= e^8.963=7808
stop posting your homework under multiple names....
To determine how much should be invested today, we can use the compound interest formula:
A = P(1 + r/n)^(n*t)
Where:
A = the future value of the investment ($12,000 in this case)
P = the principal amount (the amount to be invested)
r = the annual interest rate (6.15% or 0.0615 as a decimal)
n = the number of times interest is compounded per year (monthly compounding, so n = 12)
t = the number of years (7 years in this case)
We can rearrange the formula to solve for the principal amount (P):
P = A / (1 + r/n)^(n*t)
Now, let's plug in the values and calculate the result:
P = 12000 / (1 + 0.0615/12)^(12*7)
P = 12000 / (1 + 0.005125)^(84)
P = 12000 / (1.005125)^(84)
Using a calculator or spreadsheet software, we can calculate that (1.005125)^(84) ≈ 1.4792.
P = 12000 / 1.4792
P ≈ $8,110.50
Therefore, you would need to invest approximately $8,110.50 today at a 6.15% annual interest rate compounded monthly to have $12,000 in 7 years.