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.
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.