### 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=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 (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=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.

Return to Logarithms and modelling.