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. A=A0ektwhere k < 0.
That is, the amount of radioactive material A
present at time t is given by the formula
substance is often described in terms of its half-life, which is the time
required for half the material to decompose.
ProblemAfter 500 years, a sample of radium-226 has decayed to 80.4\% of
its original mass. Find the half-life of radium-226.
SolutionLet 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,
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. So the equation
ln ( 0.804) = 500k
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.
(1/2)A0=A0 e-0.000436 t
Now, we can finally find the half-life. We set A=1/2 A0 and
solve for t.
Dividing through by A0 again, we get:
1/2 = e-0.000436 t.
To solve for t, take natural
ln(1/2) = ln[e-0.000436 t].
Then applying the
cancellation property for logarithms yields
ln (1/2) = -0.000436 t
t= ln(1/2) /(-0.000436)
or t = 1590. The half-life is
approximately 1590 years.