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The question is:

"Show how the nonlinear regression equation y=aX^B can be converted to a linear regression equation solvable by the method of least squares."

I found how to take Y=Ae^(bX)u to a linear equation: y=a+bX+v where y=ln(Y);a=ln(A);v=ln(u).

However, I feel like there is a key difference between that example and my equation. The base of the exponential equation is e in their example and X in mine. The base being e in their example is what allows then to simplify ln(e^(bX))=bX... right? So I'm not sure what to do with my equation. I'm assuming the answer is not as simple as y=a+Bln(X).... Because this is still not linear.

Any help explaining this would be greatly appreciated!

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If $y = aX^B$ then $\log y = (\log a) + B\log X$. The intercept is $\log a$ and the slope is $B$; the independent and dependent variables are $\log X$ and $\log y$. The base of the logarithms can be any positive number except $1$.

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  • $\begingroup$ And you can solve "log(y) = log(a)+Blog(X) by the method of least squares? This is the linear equation that they are looking for? $\endgroup$ – kralco626 Nov 13 '13 at 19:07
  • $\begingroup$ It can be done if you have data. Whether it's appropriate depends on the context. It's often appropriate when you see a graph that looks like the one on the second page of this document: biostat.jhsph.edu/courses/bio621/misc/… $\endgroup$ – Michael Hardy Nov 13 '13 at 19:53
  • $\begingroup$ OK, that graph seems to jive with what we are working on in class. Thanks for the explanation. $\endgroup$ – kralco626 Nov 13 '13 at 21:00

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