If the data (Y and X) is to be fit by a function of the type $f(x)=e^{ax}$ and I fit the Y vs. log(X) via f, and obtain the fit parameter $a$, is this $a$ parameter different than the desired one (in f(x)) even though i fit Y vs. log(X) instead of Y vs. X?

Or do I have to transform the parameter somehow?

• Do you mean $y=e^{ax}+\epsilon$ and $\log(y)$ vs. $x$? – GeoMatt22 Aug 30 '16 at 0:05
• Can you write out the equation you are fitting? And also the method you use to fit (e.g. least squares)? Please put these in the question, rather than a new comment. – GeoMatt22 Aug 30 '16 at 0:18
• But the equation im fitting is given in the post, it is that of f(x) – user929304 Aug 30 '16 at 0:21
• So you are fitting $f(y)\approx x$? Depending on the error model and fitting method, the estimate for $a$ could be biased, but the "units" should be OK. – GeoMatt22 Aug 30 '16 at 0:30

Unless the data fit the equation perfectly, then the results will not be the same.

If you have a linear (Gaussian) model and specify

$Y_i = e^{ax_i} + \epsilon_i; \epsilon_i \sim N(0,\sigma^2)$

then this is not the same as saying

$\log(Y_i) = ax_i + \epsilon_i$.

As for which is correct, it depends on the situation. If the errors are all likely to have the same variance before taking logs then the first form is correct. If the errors are multiplicative then the second is appropriate. Alternatively, the errors might be doing something else and then neither approach works.

If you find the best linear fit between $Y$ and $\log_e(X)$, its slope will corresponds exactly to the parameter $a$ in your function.

Note that this only works if you're taking the log base $e$, otherwise you'd have to correct for the change of base.

• @user929304 that's what this answer is saying. In fact you would have to transform your data (or use a more sophisticated nonlinear optimization tool) if you didn't log-transform $X$ – shadowtalker Aug 30 '16 at 0:25