### Modeling Exponential Decay - Using Logarithms

A common example of exponential decay is *radioactive decay*.
Radioactive materials, and some other substances, decompose according to a
formula for exponential decay.

That is, the amount of radioactive material A
present at time t is given by the formula

A=A_{0}e^{kt}where k < 0.

A radioactive
substance is often described in terms of its **half-life**, which is the time
required for half the material to decompose.

#### Problem

After 500 years, a sample of radium-226 has decayed to 80.4\% of
its original mass. Find the half-life of radium-226.
#### Solution

Let A= the mass of radium present at time t (t=0 corresponds to
500 years ago). We want to know for what time t is A = (1/2)A_{0}.
However, we do not even know what k is yet. Once we know what k is, we can set A
in the formula for exponential decay to be equal to (1/2)A_{0}, and then
solve for t.
First we must determine k. We are given that after 500 years, the amount
present is 80.4% of its original mass. That is, when t=500, A=0.804
A_{0}. Substituting these values into the formula for exponential decay,
we obtain:

0.804
A_{0}=A_{0}e^{k(500)}.

Dividing through by
A_{0} gives us
0.804 = e^{500k}

which is an exponential equation.
To solve this equation, we take natural logs (ie. ln) of both sides. (Common
logs could be used as well.)

ln ( 0.804) = ln (e^{500k})

We know that ln
(e^{500k}) = 500k by the cancellation properties of ln and e. So the equation
becomes
ln ( 0.804) = 500k

and
k= (ln 0.804)/500.

This is the exact solution; evaluate the
natural log with a calculator to get the decimal approximation k = -0.000436 .
Since we now know k, we can write the formula (function) for the amount of
radium present at time t as

A=A_{0} e^{-0.000436 t}.

Now, we can finally find the half-life. We set A=1/2 A_{0} and
solve for t.

(1/2)A_{0}=A_{0} e^{-0.000436 t}
Dividing through by A_{0} again, we get:

1/2 = e^{-0.000436 t}.

To solve for t, take natural
logs:
ln(1/2) = ln[e^{-0.000436 t}].

Then applying the
cancellation property for logarithms yields
ln (1/2) = -0.000436 t

So
t= ln(1/2) /(-0.000436)

or t = 1590. The half-life is
approximately 1590 years.