I have a model that is roughly: $y \approx \beta_0 + \beta_1^x$.

I'm trying to figure out if there is a transformation I can do on either y or x to make it linear.

The only thing I can think of is making $x' = ln(\beta_1^x)$ -- but that's not a transformation on x solely... eg (if I don't know $\beta_1$ ahead of time, I'm not sure how to proceed...


  • $\begingroup$ Are $y$ and $x$ centered? We need further details on the variables. $\endgroup$
    – Firebug
    Commented Apr 27, 2017 at 21:29
  • $\begingroup$ @Firebug, does it matter. $\endgroup$ Commented Apr 27, 2017 at 21:29
  • $\begingroup$ Of course it does. $\endgroup$
    – Firebug
    Commented Apr 27, 2017 at 21:30
  • 3
    $\begingroup$ It's a mathematical fact, relatively straightforward to prove, that there exists no differentiable monotonic transformation of $y$ that will linearize this formula, regardless of how $x$ is transformed. Therefore any hope will rest on finding an approximately linearizing transform for the specific data you have. But why do you need to linearize? That's going to alter the implicit probability model for the errors and, besides, this is an easy model to fit: it's identical to $y = \beta_0 + \exp(\kappa x)$, an (unscaled!) exponential plus a constant. $\endgroup$
    – whuber
    Commented Apr 27, 2017 at 22:14
  • 1
    $\begingroup$ @whuber: that was my suspicion as I couldn't find one-- but I was hoping to see a proof. The request came from a a non-stats colleague for his drug toxicity model. I $\endgroup$ Commented Apr 27, 2017 at 23:05


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.