# Applications of logarithms (§§6.7&6.8)

There are several applications of exponential functions.
To solve for the input of one of these functions
is then an application of logarithms.
## Compound interest

If you invest (or borrow) an amount of money at a fixed rate of interest,
then the amount of money that you have (or owe) at the end of a period of time
is an exponential function of time.
There are three basic formulas that you want to use:
- Simple interest:
*A* = *P*(1 + *r**t*);
- Intermittent compound interest:
*A* =
*P*(1 + *r*/*n*)^{nt};
- Continuous compound interest:
*A* = *P* e^{rt}.

In these formulas, the variables have the following meaning:
*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.

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).

## Growth and decay

Anything that follows an exponential law of growth or decay
looks like
where *k* is the **relative growth rate**,
a term that can be explained using calculus.
However, 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

where *P* is the original size.
If instead the quantity goes to half its size every *h* years,
then its size after *t* years
is
(In both of these formulas, you can use different units of time than years,
as long as you do so for both *t* and *H* or *h*.)
In these formulas,
*H* is called the **doubling time**,
and *h* is called the **halflife**.
## Variations

If an object is placed in an environment at constant temperature,
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* will undergo exponential decay:
where *T* is the temperature of the environment.
You may prefer to solve for *A*:
The book's version of this formula uses different variable names,
but it is essentially the same.
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:

*A*/(*C* − *A*) =
*P*/(*C* − *P*) ⋅
e^{kt},

where *C* is the carrying capacity.
This looks rather different if we solve for *A*:
*A* =
*C**P*e^{kt} ÷
(*C* − *P* + *P*e^{kt}).

The book has a slightly different formula,
using not only different variable names
but also constants with a different meaning,
and dividing both sides of the fraction by e^{kt}
(which makes it easier to solve for *t*).

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last edited on 2015 August 31.
Toby reserves no legal rights to it.
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