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 $h=j^2 f$.
Multiply the first equation by $x^2$ and substitute $xy$ from the second:
$$x^4-x^2y^2-3x^2=x^4-4-3x^2=0.$$
This is a biquadratic equation, which gives the roots $x^2=4$ and $x^2=-1$. If you are only interested in the real solutions,
$$x=\pm2,y=\frac2x.$$
Best Answer
From the first equation we get $$z=\frac{x+2y}{4},$$ subsituting that into equation two we get $$3\Big(\frac{x+2y}{4}\Big)^2=\frac{1}{2}x^2+y^2,$$ or equivalently $$\frac{3x^2+12yx+12y^2}{16}=\frac{1}{2}x^2+y^2\Rightarrow 3x^2+12yx+12y^2=8x^2+16y^2,$$ which simplifies to $$5x^2-12xy+4y^2=0.$$ This is a quadratic in $x$ and this factorises to (by inspection or, if you like, using the quadratic formula) $$(5x-2y)(x-2y)=0\Rightarrow x=\frac{2y}{5},\quad x=2y.$$ Now, from equation one, observe that $x=4z-2y$, so $$\frac{2y}{5}=4z-2y\Rightarrow\frac{12y}{5}=4z\Rightarrow y=\frac{5z}{3}.$$ Then $$x=4z-2\Big(\frac{5z}{3}\Big)=4z-\frac{10z}{3}=\frac{2z}{3}.$$