when dealing with continuously compounded question, how do you know when to use the e function and when to use the ln(S1/S2) formula? An example would be a question such as the below:
An investor invested $10,000 into an account five years ago. Today, the account value is $18,682. What is the investor’s annual rate of return on a continuously compounded basis?
I think it’s something like 18682/10000-1 = .8682
= ln (1 + .8682) / 5
so when would you use the e (2.71828) function to solve for continuous compounding?
Just like addition and subtraction or multiplication and division, ln and e are opposites.
FV = Pe^(rt) (FV/P) = e^(rt)
Take the ln of both sides to bring the exponent down: ln(FV/P) = (rt)ln e [ln e just equals 1 and ln(FV/P) is ln(S1/S2) in this case]
Solve for r:
[ln(FV/P)]/t = r
In this case:
$18682 = $10000e^(r*5) $18682 / $10000 = e^(r*5) ln($18682 / $10000) = (r*5) ln e [ln($18682 / $10000)] / 5 = r r = 12.499508%
EDIT: you’re just rearranging the formula between ln and e.
When you’re not given dollar amounts
If you’re trying to solve for the interest rate or the investment period, you’ll get an expression using ln(St/S0); if you’re trying to solve for St or S0, then you’ll have some expression using e.
^^^^ +1 Breadmaker! Jackie Puppet
If stated rate is 8% p.a how much is contunouus rate ?
(1+i) = e^r r = ln(1.08) will give you the continuous rate.