# How to linearize a power function $y=a\cdot x^{b} + c$

I have this power function: $$y=a\cdot x^{b} + c$$ 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!

• What makes you think this is possible to do? May 20, 2021 at 12:40
• 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? May 20, 2021 at 12:42
• 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 :) May 20, 2021 at 12:54

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.
• In other words, $\log(y-c)=\log a+b\log x$, a linear relation between $\log x$ and $\log(y-c)$. May 20, 2021 at 13:10