Suppose a sample of a certain substance decayed to $65.2\%$ of its original amount after $300$ days.
What is the half-life (in days) of this substance? (Round your answers to two decimal places.)
I just cant process how to get all the values for the Decay Rate Formula: $M(t) = M(0)e^{kt}$
On
Since the figure after $300$ years is given in decimal fractions, the initial amount is implicitly given as $1$, and is not even needed for a solution.
Also, $M(t) = M(0)e^{kt}$ came in by solving a differential equation for continuous growth/decay, and isn't really needed for simpler problems like this where an yearly growth/decay rate suffices
Let the decay rate/yr be $d,\;$ and $t$ years the half-life, then
$$d^t = 0.5 \tag1$$ $$d^{300} = 0.652 \tag2$$
Dividing equation $(1)$ by equation $(2)$, taking logs and simplifying,
$t = 300\cdot\frac {log(0.5)}{log(0.652)} \approx 486.18$ years
You can not get the Initial Amount.
Luckily , you do not require the Initial amount.
SOLUTION 1 :
$M(t) = M(0)e^{kt} \tag{1}$
$M(t)/M(0) = e^{kt} \tag{2}$
We are given that $M(300)$ is $65.2 \%$ of $M(0)$
Hence , $M(300)/M(0)=65.2/100$
Plug that in (2) to get $k$ via $\log()$ :
$0.652 = e^{k \times 300} \tag{3}$
$\log (0.652) = k \times 300 \tag{4}$
You should get $k \approx -0.0014257$
Calculate $t$ with $M(t)/M(0)=50\%$ using $50\%=0.5$ & the other known values.
$0.50 = e^{-0.0014257 \times t} \tag{5}$
$\log (0.50) = -0.0014257 \times t \tag{6}$
With that , you will get half-life $t \approx 486.18$ in Days.
SOLUTION 2 :
Alternatively , you can take Initial Amount = $100$ , then at $t=300$ , we have $65.2$ , then we want $t$ where $M(t)=50$
It will give same answer.