# How to add logistic term to numeric series?

I have found a formula to calculate the number of items y after a certain value x (e.g. time) given the size of items at the beginning of the series: $$y_x = a^x y_0$$ I would like to convert it to include a logistic term 1-x/k and I know that: $$y_{x+1} = a^2 y_ x \left(1-\frac{y_x}{k} \right)$$ and $$y_{x+1} = y_x \exp \left[a(1-\frac{y_x}{k}) \right]$$ How can I convert yₓ₊₁ ∝ yₓ to yₓ ∝ y₀ including the logistic term?

How can I solve for $$y_0$$?

Thank you

• If $y_x = a^x y_0$ then (a) $y_{x+1}=a y_x$, and (b) this is exponential, not logistic, and (c) if $a > 1$ then it faces unbounded growth while if $0 < a < 1$ then it faces exponential decay Dec 22, 2020 at 12:37
• Thank you, but it is possible to convert the last equation into a y₀-equivalent? (I have no background in math so I am a little navigating in the dark...) Dec 22, 2020 at 12:43