# Log linearization

I've got the following equation:

T = T0 + K* [(gammma)^n]

I've used LOG to linearize this like: log(T-T0) = log(K) + n*[log(gamma)]

Is that ok to call log(T-T0) = y , log(K) = b , n*[log(gamma)] = a*x

Solve it (y = b + a*x) by means of a regression and then estimate T0 ?

Or would it be of so much error ?

ps: I know T0 might be between 1 and 2.

• Welcome to MSE. Please use MathJax. – José Carlos Santos Oct 30 '17 at 16:24
• As you don't know $T_0$ how do you compute the $y$'s? A low tech solution is take your best guess at $T_0$ and fit to $b$ and $a$. Use this to make a better guess at $T_0$ and then refit to $b$ and $a$ and so on until it converges to a stable value (if ever!). There are more sophisticated methods available but maybe this is good enough. – user121049 Oct 30 '17 at 17:36
• I'm gonna learn how to use this, Sr Santos. I promisse. I'm using the eye method to estimate T0. Trust me, there's no other way available and this is the point of the entire research. Cheers ! – Gilmar Neves Dec 19 '17 at 17:58

Assuming that you have $n$ data points$(\gamma_i,T_i)$ and that you want to fit the model $$T=T_0+K \gamma^n$$, remember that what you want is to minimize $$SSQ=\sum_{i=1}^n\left(T_0+K \gamma_i^n -T_i\right)^2$$ So, there are two possible ways to do it.

First way

As you thought about it, fix $T_0$ at a given value and define $y_i=\log(T_i-T_0)$ which makes the model to be linear $$y=\log(K)+n\log(\gamma)=a+n\log(\gamma)$$ and a linear regression will give you $a$ (then $K=e^a$) and $n$. When this is done, compute the predicted values for the $T$'s and compute $SSQ$ which is a function of $T_0$. Try a few values and plot the results; when you notice a minimum of $SSQ$, you have all elements to start a nonlinear regression with the real model.

Second way

Guess a value for $n$ and define $z_i=\gamma_i^n$. This makes again a linear model $$T=T_0+K z$$ and, again, a linear regression will give you the parameters $T_0$ and $K$. Repeat the same process changing $n$ until you see a minimum of $SSQ$. At this point, you have all elements to start a nonlinear regression with the real model.

The thing which is important to remember is that what is measured is $T$ and not any of its possible transforms.

• Thank you! That's the way I'm doing now (the first one). Since I have no T0 to begin with, I'm performing some estimatives based on an old method (looking at the graph and guessing the interception). I know it's terrible wrong. And for some data I use a previous linear regression and have the intercept value as T0. – Gilmar Neves Dec 19 '17 at 18:00