In a previous blog, I described how the half life of a particle can be used to solve exponential decay problems. In this blog, I will use logarithms to describe how the half life itself can be determined.
Let’s say that in 2020, I had 1 kg of a substance called Substance A. Now assume that, in 2024, 0.995 kg of that substance remains. What is the half life of Substance A?
Like before, I can let \( f(t) \) denote the mass of the substance I had at time \( t \). Moreover, let’s assume that in 2020, \( t = 0 \). Then, in 2024, \( t = 4 \). Since \( 0.995 kg \) of that substance remained in 2024,
$$ f(4) = 0.995 $$
As described my previous blog, \( f(t) \) can be written like so:
$$ f(t) = (1/2)^{t/h} $$
where \( h \) is the half life of the particle. Combining the above two equations, together, I obtain
$$ (1/2)^{4/h} = f(4) = 0.995 $$
So to determine the half live, we should solve the following for \( h \)
$$ (1/2)^{4/h} = 0.995 $$
First take logarithms on both sides, but ensure that the base of the logarithm is \( 1/2 \):
$$ \log_{1/2}((1/2)^{4/h}) = \log_{1/2} 0.995 $$
The left-hand side reduces to \( 4/h \) since
$$ \log_{1/2}((1/2)^{4/h}) = 4/h $$
(that is because, in general, \( \log_a (a^x) = x \)).
The right hand side can be written as
$$ \log_{1/2} 0.995 = \frac{\ln 0.995}{\ln 1/2} $$
(that is because, in general, \( \log_a x = \frac{\ln x}{\ln a} \)).
Putting this together, I get:
$$ 4/h = \frac{\ln 0.995}{\ln 1/2} $$
First multiply both sides by \( h \),
$$ 4 = \frac{\ln 0.995}{\ln 1/2} \cdot h $$
Then, divide both sides by \( \frac{\ln 0.995}{\ln 1/2} \). Rearranging the result, I obtain
$$ h = \frac{4 \cdot \ln 1/2}{ \ln 0.995} $$
Hence,
$$ h = 553.1302 \dots $$
This is approximately \( 553 \). So, the half life of Substance A is around five hundred and fifty three years.
