Half life of a radioactive sample

Question

The half of thallium-$201$ is $73$ hrs. How many hours will it take for an amount of thallium-$201$ to decay so that only $5%$ of the original amount remains ?

So I decide to use this

$$0.05P=P(0.05)^{73}$$

Got confused

Answer 1

A half life of $73$ hours means that after $t$ hours, what remains is $$ 0.5^{t/73} $$ of the original substance. (Note how, for $t=73$, the exponent becomes $1$, so we are left with $0.5$ of the original substance, exactly like we are supposed to. This is a quick check to make sure we have the right expression.)

Now you want to find the value of $t$ for which $5\%$ of the original substance is left. That means that you want to solve $$ 0.05=0.5^{t/73} $$

Answer 2

Here's an answer that doesn't use $e$.

The fraction you have after $t$ years is $$ f(t) = \left( \frac{1}{2} \right)^{t/73}. $$ Solve $$ f(t) = 0.05 $$ for $t$, using logarithms or by numerical trial and error.

You can do a quick estimate by noting that $$ \frac{1}{2^4} = \frac{1}{16} > \frac{1}{20} > \frac{1}{32} = \frac{1}{2^5} $$ so the answer will be beteen $4$ and $5$ periods of length $73$, probably nearer to the former - somewhat more than $300$ years.