$$g = af^b + c$$
$$i = ah^b + c$$
$$k = aj^b + c$$
I want to solve for $a$, $b$, and $c$.
$f$, $g$, $h$, $i$, $j$, and $k$ are inputs to the equations, so they don't have to be solved for. Just imagine that these variables (that are not $a$, $b$, or $c$) are numbers and are already solved for.
Basically, $a$, $b$, and $c$ must equate to a bunch of $f$'s, $j$'s, $i$'s, etc, and then that's when the problem is ready to be considered "solved".
I am already able to solve for $a$, $b$, and $c$ in a system of quadratic equations, but I am unable to solve it when it involves an unknown exponent.
There must be a way to do this because there are three equations for the three unknown variables. But I've tried everything I know about solving systems of equations and it doesn't seem to go anywhere. Thanks in advance!
Best Answer
If I properly understood the whole question, you have data points $(x_i,y_i)$ and you want to find the best fit for a model $$y=a x^b+c$$ which is intrinsically nonlinear (because of parameter $b$). The typical issue is to generate "reasonable" estimates of parameters $a,b,c$ for starting the nonlinear regression.
One thing you can notice is that the problem is very simple if $b$ is fixed. So, what you can do is to consider the function $SSQ(b)$ ($SSQ$ standing for the sum of squares of the residuals) and to plot it for various values of $b$; this just involves a linear regression. $SSQ(b)$ will go through a minimum and here you have your starting values.
For illustration purposes, I used the data given in the link provided by James Harrison. The minimum of $SS(b)$ is close to $b\approx 0.4$ and for this value $a\approx-11$, $c\approx 78$. This is more than sufficient to start the full nonlinear regression and get $$y= 83.5733 -14.5781 x^{0.35447}$$
By the way, working your three equations, suceesive eliminations of $c$ and $a$ let you with a single equation in $b$ $$F(b)=f^b (i-k)+h^b (k-g)+j^b (g-i)=0$$ to be looked graphically and may be solved by Newton method starting from a "reasonable" guess according to $$b_{n+1}=b_n-\frac{f^{b_n} (i-k)+h^{b_n} (k-g)+j^{b_n} (g-i)}{f^{b_n} \log (f) (i-k)+h^{b_n} (k-g) \log (h)+j^{b_n} (g-i) \log (j)}$$ which can be made much more compact defining a few intermediate constants.
You can also notice that writing $$F(b)=(i-k)e^{b\log(f)}+ (k-g)e^{b\log(h)}+ (g-i)e^{b\log(j)}=0$$ $$F(b)=(i-k)+ (k-g)e^{b\log(h/f)}+ (g-i)e^{b\log(j/f)}=0$$ reduces to a quadratic equation if the points, used for estimation of $a,b,c$ are selected in such a way that $\log(h/f)=2\log(j/f)$ that is to say if your three points are such that $h=j^2 f$.