Trying to solve this question:
A radioactive material is known to decay at a yearly rate proportional to the amount at each moment. There were 2000 grams of the material 10 years ago. There are 1990 grams right now. What is the half-life of the material?
I've set it up as $p = p_0 * e^{kt} \implies 1990 = 2000 * e^{k(10)}$, then solve for k, but that doesn't seem to work.
The formula they give below seem to imply I should use negative k, which I suppose I can reconcile since it's a decay instead of growth.
However I cannot understand why they would:
- replace e with 2..? (Where did the e go..?)
- Assume H = time, why they would divide it by time..?
Best Answer
We have a "formula" for radioactive decay which is $p(t) = p_0e^{-kt}$ where $p_0 > 0$, $k> 0$ and $t > 0$. So in this case, we know the following: $p_0 = 2000$ and $p(10) = 1990 = 2000e^{-10k}$. This leads to $k = \frac{\ln\left(\frac{1990}{2000}\right)}{-10} \approx 0.000501254182354$.
But more importantly, we need to transform this into $p = p_0\left(\frac{1}{2}\right)^{\frac{t}{k}}$, then whenever $t = k$ is a half-life. So now the real formula presents itself:
$$ 1990 = 2000\left(\frac{1}{2}\right)^{\frac{10}{k}} $$
Then we solve:
$$ \frac{10}{k} = \frac{\log\left(\frac{1990}{2000}\right)}{\log\left(\frac{1}{2}\right)} \leadsto k = 10\frac{\log(2)}{\log\left(\frac{2000}{1990}\right)} \approx 1382 $$
A word about $r^\frac{t}{k}$
You could just have easily written $\left(\frac{1}{2}\right)^{\frac{t}{k}} = 2^{-\frac{t}{k}}$. It's probably even more canonical to write $2^{\pm kt} = 2^{\pm \frac{1}{\lambda}t}$. And that gets to the point of writing it one way or the other: for $2^{kt}$, $k$ is a "frequency" (that's hard to relate within a real exponential) and $\lambda = \frac{1}{k}$ is a "wavelength"--a distance--in this case a period of time: the half life...or third life...or three quarters life or even the exotic $\frac{1}{e}$ life--it's whatever base you decide you want:
$$ p_n = p_0\cdot \left(\frac{1}{b}\right)^{\frac{1}{\lambda}\cdot\{t = n\}} \leadsto \lambda = n\frac{\log(b)}{\log\left(\frac{p_0}{p_n}\right)}\text{, where }b > 1, n > 0, p_0 > p_n $$
Note that in this formula, the assumption is that it's a decay, i.e. that this can be written as $p_0b^{-kt}$ where $b > 1$ and $k > 0$. A negative exponential defines a decay! But we can rewrite knowing what a negative exponential means: $b^{-kt} = \frac{1}{b^{kt}} = \left(\frac{1}{b}\right)^{kt}$. Then, by rewording to $k = \frac{1}{\lambda}$ we can ask what happens when $t = \lambda$, i.e. the exponent equals $1$.