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 $y=\ln(Y)$, $a=\ln(A)$ and $v=\ln(u)$.
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 $y=a+B \ln(X)$. Because this is still not linear.
Any help explaining this would be greatly appreciated!
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 $1$.