# Linear regression method help?

Let's suppose we have a function $Y=A\cdot t^B$ and the values for $Y$ are $30,60,90,120,150$ and the values for $t$ are respectively $0.974, 1.331, 1.718, 1.971, 2.356$. Can you find $A$ and $B$ with the method of linear regression? I have to do a lab work and this is a very small part of it, which does not count but I still have to do it and I have never done linear regression, I need this now? please?

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can't you find A and B without regression? – dato datuashvili Jan 4 '13 at 10:17
I need that with the regression method..that's what it says in my paper – ddsd Jan 4 '13 at 10:18
then i think it is not linear regression model,it is called exponential regression – dato datuashvili Jan 4 '13 at 10:19
in case of any question,please ask – dato datuashvili Jan 4 '13 at 10:30
I highly doubt that the pedagogical purpose of this question is to do power/exponential regression. A straight line can be fitted using linear least squares fit. For power/exponential regression, one has to use non-linear least squares fitting...which is much more non-trivial than LLSF. – Fixed Point Jan 4 '13 at 10:55

Are you allowed to use a calculator or software or do you have to do this by hand? In either case, if you have $y=at^b$ then just take the logarithm of both sides and get $$\log(y)=\log(a)+b\log(t)$$ and then use linear regression using this form of the fitting function. So remember you won't put in your data points $y$ and $t$ as you have them above. You have to take their log and use $\log(y)$ and $\log(t)$. Then the slope of the fitted line will be your $b$ because that is the coefficient in front of $\log(t)$ and the y-intercept will be $\log(a)$ so exponentiate it to get $a$ and VOILA!
Yep, they wouldn't be exactly the same but they are close and in this case both answers are within 95% confidence interval of each other. The power regression gives me $a=36.45$ and $b=1.679$. – Fixed Point Jan 4 '13 at 11:18