### Modelling 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=A0ekt
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)A0. 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)A0, 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 A0. Substituting these values into the formula for exponential decay, we obtain:

0.804 A0=A0ek(500).

Dividing through by A0 gives us
0.804 = e500k

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 (e500k)

We know that ln (e500k) = 500k by the cancellation properties of ln and e (see Logarithms). 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=A0 e-0.000436 t.

Now, we can finally find the half-life. We set A=1/2 A0 and solve for t.

(1/2)A0=A0 e-0.000436 t

Dividing through by A0 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.