How to contextualize an inverse function for interpolation purposes? I have a function that describes the growth of a population in a logistic term:
$$N_t = rN_0(1-\frac{N_0}{\kappa})$$
I would like to interpolate the values so that, given a known $y$ I can get the related $x$. To make it easier, I linearized the equation by taking log($N_t$). I removed the point that were not in a straight line and the obtained a linear model:
$$log(N_t) = y = st + i = mx + b$$
Using R, I can interpolate x given the model y~x , as in this picture:

But what I need to do is the opposite:

Since in R is not possible to get y from the model, and it is not correct to invert the model as x~y, I found the inverse function as:
$$log(y) = sx+i$$ (where $s$ and $i$ are the slope and intercept of the first model
$$log(x) = sy+i$$
$$y = \frac{log(x) -i}{s}$$
But now I am lost on what to do next. Shall I use the inversed function on the original x or y values? How do I convert these values into the original space's one? In other words, I get something like this:

How do I use the inverted function?
Manually, I have found that if $y=1000000$, then $x=16.6$. The parameters are: $s=0.022$, $i=0.328$. If I solve, I get:
$$log(x) = s \cdot 10^6 -i$$
$$x = 10^{s \cdot 10^6 -i}= ∞$$ OR
$$x = 10^{s \cdot 6 -i}= 0.6$$
My apologies but I am not a mathematician.
Thank you
 A: I think there's some confusion around the meaning of the first equation. Let me write it this way
$$
N_1 = r N_0\left(1 - \frac{N_0}{\kappa} \right) \tag{1}
$$
this means if you know the population at time $t=0$ you can calculate the population at $t=1$ and that's it. So the question is, how do you calculate the value of the population at $t=2$? something very similar to Eq. (1)
$$
N_2 = r N_1\left(1 - \frac{N_1}{\kappa} \right) \tag{2}
$$
where the value of $N_1$ is already known. The pattern is: if I want to know the value of the population at instant $t$ I need to know its value at $t-1$
$$
N_{t} = r N_{t-1}\left(1 - \frac{N_{t-1}}{\kappa} \right) ~~~t=1,2,\cdots\tag{3}
$$
This is a recursive expression, to get the value $N_t$, you need to calculate all the preceding values, namely $N_{t - 1}, N_{t-2}, \cdots, N_1, N_0$, Here's an example of for $r=3$ and $\kappa=1$
$$\begin{array}{cc} \hline
t & N_t \\ \hline
0 & 0.4 \\
1 & 0.792 \\
2 & 0.543628 \\
3 & 0.818718 \\
4 & 0.489781 \\
5 & 0.824655 \\
\vdots & \\
995 & 0.479427 \\
996 & 0.823603 \\
997 & 0.479427 \\
998 & 0.823603 \\
999 & 0.479427 \\ \hline
\end{array}
$$
or in graphical form

From this, you immediately can see that your original question may not be solved, e.g. what is the value of $t$ for which $N_t = 0.479427$?

Additional comments


*

*Here's another example for a different value of $r$

this is a textbook example of chaotic behavior, so in that regard, your question is similar to ask "which will be the times at which the temperature outside is 27 C?"
A: since $a_n$ convergent to $a$, we multiple $a_n$ to your sequence and plus an extra term $b_n^{2}/2$, ignore whether $b_n$ is convergent or not at first.
then product your sequence to be $\frac{(aa_n)^2-2aa_nb_n+b_n^{2}}{2}-b_n^{2}/2=(aa_n-b_n)^2/2-b_n^{2}/2$. since $a\in{(0,1]}$, so $aa_n\in{(0,1]}$ be bounded (avoid squares ?), the above terms can be merge to $aa_n$ and $aa_n-2b_n$, only depends one term for $b_n$. now your conclusion seems reliable.
