- Simple interest:
*A*=*P*(1 +*r**t*); - Intermittent compound interest:
*A*=*P*(1 +*r*/*n*)^{nt}; - Continuous compound interest:
*A*=*P*e^{rt}.

*P*is the original amount of money, called the**principal**;*A*is the amount of money*after*a period of time;*t*is the length of time (in years);*r*is the (annual) rate of interest; and*n*is the frequency (the number of times per year) that interest is compounded.

With simple interest,
the interest is applied once, at the end of the time period;
this is effectively compound interest where *n* = 1/*t*.
With intermittent compound interest,
the interest is applied several times and added to the original amount,
so that interest can be charged on the interest later.
With continuous compound interest,
the interest is added to the original amount continuously;
this is like compound interest where *n* is effectively infinite.

Here is an example,
which will also illustrate how continuous compound interest comes about:

Or download it:
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Another way to see where the special number e comes into it
is to use some properties of exponents
to write the formula for intermittent compound interest
as *A* =
*P*(1 + *r*/*n*)^{nt} =
*P*(1 +
*r*/*n*)^{(n/r)(rt)} =
*P*((1 +
*r*/*n*)^{n/r})^{rt} =
*P*((1 + *x*)^{1/x})^{rt},
where I've written *x* for *r*/*n*
(so that *n*/*r* is its reciprocal, 1/*x*).
With *r* fixed and positive (as 0.06 for example),
as *n* gets arbitrarily large,
*x* = *r*/*n*
will get arbitrary close to 0 while remaining positive.
So if (1 + *x*)^{1/x}
gets arbitrarily close to some number e
as *x* gets arbitrarily close to 0 while remaining positive,
then *A* gets arbitrarily close to *P*e^{rt}
as *n* gets arbitrarily large.
(In Calculus, this sort of thing is called a *limit*.)
And that's exactly what happens;
you can approximate e as closely as you like
by using a sufficiently small positive number *x*
in the expression (1 + *x*)^{1/x}.
(Compare the book's definition of e on the top of page 431;
they write 1/*n* in place of *x*,
which amounts to using *r* = 1.)

If you know the final amount *A*
and want to find the principal *P*,
then solve the equation for *P*.
If you know both *A* and *P*
and want to find the amount of time *t*,
then you must take a logarithm to solve the equation.
It's also possible to solve for *r*,
but not for *n*
(at least not with the operations that we use in this class).
Here are the results
in the case of intermittent compound interest (which is the most complicated):

*P*=*A*(1 +*r*/*n*)^{−nt};*t*= log_{1+r/n}(*P*/*A*)/*n*;*r*=*n*(^{nt}√(*A*/*P*) − 1).

*A*=*A*_{0}e^{kt}.

*k*= ln(*A*/*A*_{0})/*t*;*t*= ln(*A*/*A*_{0})/*k*.

You can replace e with any other valid base (2, 10, whatever),
so long as you change *k* to match.
A different choice of the base can make the correct value of *k*
either more or less obvious.
For example, if a quantity doubles in size every *H* years,
then its size after *t* years
is

*A*=*A*_{0}2^{t/H}.

*A*=*A*_{0}2^{−t/h}.

If an object is placed in an environment at constant temperature,
then it will cool down or heat up to reach the environment's temperature.
This temperature will neither grow nor decay exponentially;
but according to Isaac Newton's **law of cooling and heating**,
the *difference in temperature* between the object and its environment
will undergo exponential decay.
If *u* is the temperature of the object
and *T* is the temperature of its environment,
then *u* − *T* is the quantity *A*
in the general formula for exponential growth and decay,
with *u*_{0} − *T*
in place of *A*_{0}:

*u*−*T*= (*u*_{0}−*T*)e^{kt}.

*u*=*T*+ (*u*_{0}−*T*)e^{kt}.

Exponential decay is one thing,
but exponential growth forever is unrealistic.
In the model of **logistic growth**,
there is a **carrying capacity**
beyond which a population cannot grow.
In this case, there is still an exponential growth,
but it is *the ratio of the population to the remaining capacity*
that grows exponentially.
If *P* is the population and *c* is its carrying capacity,
then *P*/(*c* − *P*)
is the *A* in the general formula for exponential growth and decay,
with *P*_{0}/(*c* − *P*_{0})
in place of *A*_{0}:

*P*/(*c*−*P*) =*P*_{0}/(*c*−*P*_{0}) e^{kt}.

*P*=*c**P*_{0}e^{kt}/(*P*_{0}e^{kt}+*c*−*P*_{0}).

*P*=*c*/(1 +*a*e^{−kt}).

Go back to the course homepage.

This web page and the files linked from it were written between 2012 and 2016 by Toby Bartels, last edited on 2016 September 2. Toby reserves no legal rights to them. The linked files were produced using GIMP and recordMyDesktop and converted from Ogg Vorbis to other formats by online-convert.com.

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