The question is:
"Show how the nonlinear regression equation
y=aX^Bcan 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!