How do you Find exponential decay rate?

Explanation:
Exponential decays typically start with a differential equation of the form:
${\displaystyle \frac{dN}{dt}\propto -N\left(t\right)}$
That is, the rate at which a population of something decays is directly proportional to the negative of the current population at time ${\displaystyle t}$. So we can introduce a proportionality constant:
${\displaystyle \frac{dN}{dt}=-\alpha N\left(t\right)}$
We will now solve the equation to find a function of ${\displaystyle N\left(t\right)}$
${\displaystyle \to \frac{dN}{N}=-\alpha d\left(t\right)}$
${\displaystyle \to {\int }_{\{}^{\{}\frac{dN}{N}={\int }_{\{}^{\{}-\alpha dt\to \ln \left(N\right)=-\alpha t+C}$
${\displaystyle \to N\left(t\right)=A{e}^{\alpha t}}$ where ${\displaystyle A}$ is a constant
This is the general form of the exponential decay formula and will typically have graphs that look like this:
graph ${\displaystyle \left\{e^{-x}[-1.465,3.9,-0.902,1.782]\right\}}$
Perhaps an example might help?
Consider a lump of plutonium 239 which initially has ${\displaystyle {10}^{24}}$ atoms. After one million years have elapsed years the plutonium now has ${\displaystyle 2.865\times {10}^{11}}$ atoms left. Work out, ${\displaystyle A}$ and ${\displaystyle \alpha }$ . When will the plutonium have only ${\displaystyle 5\times {10}^8}$ atoms left and what is the decay rate here?
We are told the lump has ${\displaystyle {10}^{24}}$ atoms at t = 0 so:
${\displaystyle N\left(0\right)=A{e}^0={10}^{24}\to A={10}^{24}}$
Now at 1 million years: ${\displaystyle {10}^{6}years}$
${\displaystyle N\left({10}^6\right)={10}^{24}{e}^{-\alpha \left({10}^6\right)}=2.865\times {10}^{11}}$
Rearrange to get:
${\displaystyle \alpha =-\frac{1}{{10}^6}\ln \left(\frac{2.865\times {10}^{11}}{{10}^{24}}\right)\approx 2.888\times {10}^{-5}yr\{-1\}}$
So ${\displaystyle N\left(t\right)={10}^{24}{e}^{-2.888\times {10}^{-5}t}}$
For the next part:
${\displaystyle N\left(t\right)=5\times {10}^8={10}^{24}{e}^{-2.888\times {10}^{-5}t}}$
Rearrange to get t:
${\displaystyle t=-\frac{1}{2.888\times {10}^{-5}}\ln \left(\frac{5\times {10}^8}{{10}^{24}}\right)\approx 1.22\times {10}^{6}yr}$
Now for the last part, the decay rate is already defined a way back at the very start, simply evaluate it at the given time:
${\displaystyle \frac{dN}{dt}=-\alpha t=-2.888\times {10}^{-5}(1.22\times {10}^6)}$
= -35.23 atoms per year.
The idea is to start with differential equation above, which gives the decay rate, and solve it to get the population at any given time.