What interest rate, compounded weekly, would you need to be paid in order to grow an $8000 investment into $12,000 over a ten year period?
See solution to your previous question:
http://www.jiskha.com/display.cgi?id=1291783815
To calculate the interest rate required to grow an investment from $8000 to $12,000 over a ten-year period, compounded weekly, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount (initial investment)
r = the annual interest rate (in decimal form)
n = the number of times interest is compounded per year
t = the time period in years
In this case, we know:
A = $12,000 (desired future value)
P = $8000 (initial investment)
n = 52 (compounded weekly)
t = 10 years
Now, let's plug the values into the formula and solve for r:
$12,000 = $8000(1 + r/52)^(52*10)
First, rearrange the equation:
(1 + r/52)^(52*10) = $12,000/$8000
Simplify the right side:
(1 + r/52)^(520) = 1.5
Take the logarithm of both sides:
log[(1 + r/52)^(520)] = log(1.5)
Using the logarithmic property, bring down the power:
520 * log(1 + r/52) = log(1.5)
Now, divide both sides by 520:
log(1 + r/52) = log(1.5)/520
Finally, raise both sides as exponentials of base 10:
1 + r/52 = 10^(log(1.5)/520)
Subtract 1 from both sides:
r/52 = 10^(log(1.5)/520) - 1
Multiply both sides by 52:
r = 52 * (10^(log(1.5)/520) - 1)
Using a scientific calculator or computer software, calculate the expression inside parentheses, and then multiply by 52 to obtain the required interest rate as a decimal or percentage.