How do you Find exponential decay rate?

Explanation: 
Usually, exponential decays begin with a differential equation of the following form:
${\displaystyle \frac{dN}{ dt }\propto -N\left(t\right)}$ 
In other words, the rate of population decay is directly inversely proportional to the current population at a given 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.