For a given set of data, the power form of the regression equation is given by $$y=b\cdot x^{m}$$ where $$m=\frac{n(\sum \mathrm{ln}(x_i)\mathrm{ln}(y_i))-(\sum \mathrm{ln}(x_i))(\sum \mathrm{ln}(y_i))}{n(\sum (\mathrm{ln}(x_i))^2)-(\sum \mathrm{ln}(x_i))^2}$$ $$\textrm{ln}(b)=\frac{\sum \textrm{ln}(y_i)-m(\sum \textrm{ln}(x_i))}{n}$$
However, the resulting equation $y=b\cdot x^{m}$ is centered at $x=0$. My question is, how should this method be modified so that the whole graph shift a certain $c$ units to the right along the x-axis instead of centered at $x=0$? In other words, the form of the equation should be:
$$y=b\cdot (x-c)^{m}$$
So for the data below (black dots), for example, instead of getting this (centered at $x=0$, $y\rightarrow \infty$ as $x\rightarrow 0$):
I should get a regression line like (centered at $x<0$, y-intercept exists as $x=0$):
In the first part, in order to fit the data to $y=b\, x^m$, you linearize the model to get $\log(y)=\beta=m\log(x)$ which corresponds to a linear regression. But, this gives you estimates of the parameters. From this points, you need to continue with nonlinear regression since what you want to minimize is the sum of the squared errors for the $y$'s and not for the $\log(y)$'s.
For the problem where the model is $$y=b\,(x-c)^m$$ what you could notice is that, if $c$ was known, it would be the same problem as before. So, fix $c$ to a given value, make the logarithmic transform and compute $SSQ(c)$ since the resulting values of $\beta$ and $m$ depend on $c$; plotting the function will show you an approximate value of the $c$ which minimizes $SSQ(c)$.
What this is done, you have all elements for starting the nonlinear regression with the three adjustable parameters.
Edit
For illustration purposes, I generated a table of $25$ data points $(x_i=i+20\, , i=1$ to $20)$ with $$y_i=12.34 (x-15.27)^{-0.345}+0.2 (-1)^i$$ and I did what was described above fixing successively $c$ to $0,1,\cdots,24$ (remember than $c$ cannot be larger than the smallest of the $x_i$'s because of the exponentiation. For the linearized model, the sum of squares is a quite flat function of $c$ but it goes through a minimum for $c=14$ (at this point, $\log(b)=2.637$ and $m=-0.379$). Now, we can start the nonlinear regression which leads to $b=14.392$, $c=13.692$, $m=-0.3866$.