I've been taught a very different equation to the normal logistic equation on the internet. Thing is, I don't know how to convert from one form to another so any help would be great.
The equations online is: $\frac{dp}{dt} = kP(1-\frac{p}{M})$, where $M$ is the carrying capacity.
The equation I was taught: $\frac{dp}{dt} = ap(\frac{k}{p}-p)$ where $\frac{k}{p}$ is the carrying capacity. Replacing this carrying capacity with $M$ gives:
$$\frac{dp}{dt} = ap(M-p)$$
I'm not sure how $M$ can be turned into a denominator of a fraction, such as the equation above.
For the logistic equation
$$\frac{dp}{dt}=kp(1-\frac{p}{M})$$
replacing $M\rightarrow \frac{k}{p}$ gives
$$\frac{dp}{dt}=kp\left(1-\frac{p}{\frac{k}{p}}\right)=kp\left(1-\frac{p^2}{k}\right)$$ $$=kp\left[\frac{p}{k}\left(\frac{k}{p}-p\right)\right]=p^2\left(\frac{k}{p}-p\right)=ap\left(\frac{k}{p}-p\right)$$
where $a=p$
Or the other way around setting $\frac{k}{p}\rightarrow M$ in $\frac{dp}{dt} = ap(\frac{k}{p}-p)$ gives
$$\frac{dp}{dt} = ap(M-p)=ap\left[M\left(1-\frac{p}{M}\right)\right]$$ $$=aMp\left(1-\frac{p}{M}\right)=kp\left(1-\frac{p}{M}\right)$$
where $k=aM$, which is a constant of proportionality.