# Logistic growth inflection point

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...

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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 $1/2 = y = 1/(1 + \exp(-x_0))$? –  whuber Oct 12 '12 at 14:58

Solving for x, you get:

x = (-1/b) * ln( (1-y) / (a*y) )


Insert y = 0.5 and you get:

x = (-1/b) * ln(1/a)


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

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