Expert: Paul Klarreich - 2/10/2006

Question
Sean invests $10,000 at an annual rate of 5% compounded continuously according to the formula A=Pe^(rt) where A is the amount, P is the principal, e=2.718, r is the rate of interest, and t is time, in years.
Determine, to the nearest dollar, the amount of money he will have after 2 years.
Determine how many years, to the nearest year, it will take for his initial investment to double.

Answer
Hi, Lauren,

Your Question:  Sean invests $10,000 at an annual rate of 5% compounded continuously according to the formula A=Pe^(rt) where A is the amount, P is the principal, e=2.718, r is the rate of interest, and t is time, in years.
Determine, to the nearest dollar, the amount of money he will have after 2 years.
Determine how many years, to the nearest year, it will take for his initial investment to double.
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Would you believe there was a time when banks actually did this?  They were not allowed to compete by offering higher interest rates, so instead they advertised more frequent compounding.  Instead of quarterly compounding, they claimed monthly -- when other banks caught on, it became weekly, then daily.  Finally, they hired a mathematician who explained how to do continuous compounding.  Amazing, especially since computers hadn't been invented yet.  And you probably weren't born yet, either.

Now then, about your problem.  In the 'continuous compounding' formula:

A = P exp(rt) -- that's how mathematicians write it.

A is the amount you will have, after
t years, when you put
P dollars into the bank, which pays
r as interest.

Of course the interest must be expressed as a decimal fraction.  So 5% interest is written as r = 0.05

[By the way,  e is not 2.718,  e is e, and it is built into your calculator.]

So for your first problem, set:

P = 10000
r = 0.05
t = 2

and compute  A = 10000 exp(0.05 * 2) = 10000 exp(0.1) = $11051.71

[Note: On your Windows XP calculator, you enter these items:

0.1
Inv
ln
*
10000
<enter>

The Exp button you see there does NOT do the exponential function, apparently.  

You use the fact that  exp(x) is the inverse of the natural logarithm function, written  ln(x).
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For the second problem, set:

A = 2   (never mind the 000000000's -- irrelevant)
P = 1
r = 0.05  as before

and you want to find t.

2 = exp(0.05 t)

Now 0.05 t = ln(2)  and  t = ln(2)/0.05 = 13.86  years.
Press these keys:
2
ln
/
0.05
<enter>