What interest rate, compounded monthly, would you need to be paid in order to grow a $2000 investment into $3500 over a five year period?
Use the compound interest formula for
n=number of periods = 5*12 = 60
R=monthly interest rate, 1.01 for 1%
P=present value of investment = $2000
F=future value of investment = $3500
Then
F=ARn
3500=2000R60
rearrange
Rn = 3500/2000 = 1.75
Take log on both sides
60*log(R)=log(1.75)
log(R)=log(1.75)/60
R = elog(1.75)/60
=1.00937
The equivalent annual interest. r
=(R-1)*12=11.24%
Note: in the above expression, I used log(x) for natural log, usually written as ln(x). You would have got the same answer using log to the base 10, but at the end, the values must be raised to the power of 10 (instead of e).
To calculate the interest rate needed to grow a $2000 investment into $3500 over a five-year period with monthly compounding, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment ($3500)
P = the principal or initial investment ($2000)
r = the interest rate (in decimal form)
n = the number of times that interest is compounded per year (monthly, so n = 12)
t = the number of years (5 years)
We need to solve the equation for r. Rearranging the formula, we get:
(1 + r/n)^(nt) = A/P
Now, let's plug in the given values:
(1 + r/12)^(12*5) = 3500/2000
Simplifying further:
(1 + r/12)^60 = 1.75
To find the interest rate, we need to isolate (1 + r/12)^60. We can take the 60th root of both sides of the equation:
(1 + r/12) = (1.75)^(1/60)
Calculating the right side of the equation using a calculator, we get:
(1 + r/12) ≈ 1.02329
Now, we need to isolate r by subtracting 1 from both sides:
r/12 ≈ 1.02329 - 1
r/12 ≈ 0.02329
Finally, multiply both sides by 12 to solve for r:
r ≈ 0.02329 * 12
r ≈ 0.2795
Therefore, the interest rate, compounded monthly, would need to be approximately 0.2795 or 27.95% in decimal form to grow a $2000 investment into $3500 over a five-year period.