Simply I have a function which looks like this: $$\lambda =\lambda_0 \cdot \exp(-T_0/T)\;,$$, where $\lambda_0$ and $T_0$ are unknown constants. By appplying to both sides ln() I get : $$\ln(\lambda)=\ln(\lambda_0)-T_0(1/T)$$ It's of form $y=b+a/x$. The question is how can I apply linear regression formulae? Maybe smart substitution will be enough ?

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    $\begingroup$ As long as $x$ is known, clearly also $1/x$ is known! This is just the usual linear regression with $1/x$ as precictor. Just calculate $z=1/x$ and use linear regression with $z$ as predictor. Or, if you are using R, use in the model formula $y \sim I(1/x)$. $\endgroup$ – kjetil b halvorsen Feb 20 '17 at 10:25
  • $\begingroup$ @kjetil b halvorsen thanks for claryfying the issue, what do you mean by R ? $\endgroup$ – Sebastian Pinocy Feb 20 '17 at 10:39
  • $\begingroup$ @kjetilbhalvorsen why not write in as an answer instead of comment ? $\endgroup$ – Branislav Cuchran Feb 20 '17 at 10:44
  • $\begingroup$ R is a statistical programming language / software package. See here. To apply a linear regression, you should include an additive error term in your model. $\endgroup$ – Richard Hardy Feb 20 '17 at 10:44
  • $\begingroup$ Your second model is not the same as the first: it posits a different pattern of errors. Depending on what patterns are likely to hold for your data, you might need weighted least squares, a nonlinear regression, or even something else. $\endgroup$ – whuber Feb 20 '17 at 15:48

Yes, if you take $\frac{1}{T}$ as predictor and $\ln(\lambda)$ as response, you just have a very usual linear regression problem.

Anyway, you should remember to check linear regression assumptions with the transformed variables.

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    $\begingroup$ Note that $T_0$ is unknown so you can't use it to make a predictor. (You may want to read whuber's comment under the question as well) $\endgroup$ – Glen_b Feb 21 '17 at 2:41

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