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I have a model to fit but I am not sure if it is correct:

Is $\exp(ax+bz+c)^d$ algebraically the same as $\exp(dax+dbz+dc)$?

Edit

what about this one?

Is $[exp(ax+bz+c)+j]^d$ algebraically the same as as $[exp(dax+dbz+dc)+dj]$

Where a,b,c,j,d are parameters for non-linear regression fit.

If so, why? Can you explain please?

Barry
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  • It would help to have more parentheses there: both $(\exp(x))^d$ and $\exp(x^d)$ are unambiguous, while $\exp(x)^d$ is something that will make the reader wonder about the intended meaning. –  Mar 11 '15 at 00:07
  • @Woodface Thanks my question is edited. – Barry Mar 11 '15 at 08:40

1 Answers1

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Yes, it is correct. Note that $\exp(z) = e^z$, where $e$ is a number given by $e = 2.71828...$. It then follows from a basic property of exponents that $$ \exp(ax + bz + c)^d = \\ \left[e^{ax+bz+c}\right]^d =\\ e^{(ax+bz+c)d} =\\ e^{dax + dbz + dc} =\\ \exp(dax + dbz + dc) $$

Ben Grossmann
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