I have this power function:

Power regression

$$y=a\cdot x^{b} + c$$ and I would like to linearise it into the form: $$y=a\cdot x + b$$ I am trying to linearise it because most algorithms for regression are usually done with linear regression. Colleagues of mine told me that I could perhaps put it into the log space to linearise it and do the same statistics as a linear regression. So, I have been trying to put it into the log space but I got stuck. There is ton of help online for the equation in the following form: $$y=a\cdot x^{b} $$ $$log(y) = log(a\cdot x^{b})$$ $$log(y) = b\cdot log(x) + log(a)$$ But I cannot find a way to include the $c$ into this and I need it in my case. So I am stuck here: $$log(y) = log(a\cdot x^{b} + c)$$ Any help would be very much appreciated!

  • $\begingroup$ What makes you think this is possible to do? $\endgroup$ Commented May 20, 2021 at 12:40
  • $\begingroup$ Is there a reason you think it could be done? You can solve for $x$ easily enough, so exactly why do you want to do this? $\endgroup$ Commented May 20, 2021 at 12:42
  • $\begingroup$ Hello, made a slight edit with a graphic of what I am talking about. I hope it will orientate better the reader. I am not sure it would be possible so I am giving my shot :) $\endgroup$ Commented May 20, 2021 at 12:54

1 Answer 1


I think the best you can do is to subtract $c$ from $y$ and treat $y-c$ as the quantity you are considering.

If you are trying to linearize it in order to perform least squares optimization on the data, your problem is that you have $3$ parameters instead of $2$. You can't convert it to linear because a line has only two parameters.

  • 4
    $\begingroup$ In other words, $\log(y-c)=\log a+b\log x$, a linear relation between $\log x$ and $\log(y-c)$. $\endgroup$ Commented May 20, 2021 at 13:10
  • $\begingroup$ I see, thank you for your answer! I think that I am then doomed to do the statistics on the power regression. Thanks anyway! $\endgroup$ Commented May 23, 2021 at 8:45

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