I have a logistic growth curve as follows:

$y = \frac{1}{(1 + ae^{-bx})}$, where x is the independent measure (x-axis) and a and b are paramaters. The inflection point of this equation is when y = 0.5.

Given that information, I need to find the x-value at y = 0.5. But this equation isn't easy to solve. I'm using the statistical language R to find these fits.

So, given the parameters a and b, how can I find the x-value at y = 0.5 (and again, I can't do things like derivatives because I am using a programming language). By the way, I did try to solve for x, but it's quite ugly and I'm not sure how to do it...

  • $\begingroup$ Hint: Writing the formula in the location-scale form $y = (1 + \exp(-b(x - \log(a)/b)))^{-1}$ shows that if $x_0$ is the solution when $\log(a)=0$ and $b=1$, then $x_0 + \log(a)/b$ is the general solution. Can you find $x_0$ such that $1/2 = y = 1/(1 + \exp(-x_0))$? $\endgroup$
    – whuber
    Oct 12, 2012 at 14:58

1 Answer 1


Solving for $x$, you get:

$$x = -\frac{1}{b} \ln\left(\frac{1-y}{ay}\right)$$

Insert $y = 0.5$ and you get:

$$x = -\frac{1}{b} \ln\left(\frac{1}{a}\right)$$

It isn't that ugly... Practice the math - I promise it will help you.

  • $\begingroup$ ln(-1/a)??? but "a" is positive. $\endgroup$
    – CodeGuy
    Oct 12, 2012 at 1:59
  • $\begingroup$ @CodeGuy then that point is not on the plot. Look closely: If y equals 0.5, then (1-ae^(-bx))=2, meaning ae^(-bx)=-1 right? now e^(-bx) is always positive, so a has to be negative always. $\endgroup$
    – Bitwise
    Oct 12, 2012 at 2:15
  • $\begingroup$ oh sorry...the original function should be 1 + a... $\endgroup$
    – CodeGuy
    Oct 12, 2012 at 2:36
  • $\begingroup$ @CodeGuy I changed the question and answer accordingly. $\endgroup$
    – Bitwise
    Oct 12, 2012 at 10:36

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